{"id":2140,"date":"2026-10-04T13:37:01","date_gmt":"2026-10-04T13:37:01","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=2140"},"modified":"2026-10-04T13:37:02","modified_gmt":"2026-10-04T13:37:02","slug":"right-hand-rule-physics","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/right-hand-rule-physics\/","title":{"rendered":"The Right-Hand Rule in Physics: Three Clear Rules, One Hand"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\n\nRight hand rule physics is the set of hand rules that turn a cross product into a direction. Lay the\nfingers of your right hand along the first vector, curl them towards the second, and your thumb\ngives the third: the magnetic force on a moving charge, the field circling a current, or the north\nend of a coil.\n\n<\/p><\/div>\n\n<p>Almost every direction question in magnetism arrives in the same shape. You are handed two things that point somewhere \u2014 a velocity and a field, a current and a place to stand, a current and a coil \u2014 and asked for a third that sits at right angles to both. The algebra alone cannot finish the job, because a plane has two perpendicular sides and the equations treat them identically.<\/p>\n\n<p>So fix the axes once, before a single direction word is used: x counts positive to your right across the screen, y counts positive upwards, and z counts positive straight out of the page towards your eye. Into the page is therefore negative z. Assume the opposite sense of z and every answer on this page comes out backwards, which is the single commonest way this topic goes wrong.<\/p>\n\n<h2>What Is the Right-Hand Rule in Physics?<\/h2>\n\n<p>The right-hand rule is a convention for reading the direction of a cross product off your own hand: fingers along the first vector, curled towards the second, thumb along the product.<\/p>\n\n<p>That is the whole mechanism. What makes it feel like three separate tricks is that physics keeps writing cross products with different letters in them, and each version has picked up its own name and its own folklore.<\/p>\n\n<p>It is worth being clear about what this page does and does not do. For what a magnetic field actually is, how it bends a charge onto a circle, and the whole of the F = qvB sin\u03b8 arithmetic, read our guide to the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/magnetic-field\/\">magnetic field<\/a>, which answers how big; if you already have a speed, a field and the angle between them and simply want newtons, our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/magnetic-force\">Magnetic Force Calculator<\/a> does that sum. This page answers which way.<\/p>\n\n<h3>Three rules, one piece of physics<\/h3>\n\n<ul>\n<li><strong>The force rule.<\/strong> A charge moving through a magnetic field feels a push at right angles to both its velocity and the field.<\/li>\n<li><strong>The grip rule.<\/strong> A current in a straight wire wraps a magnetic field round itself in circles.<\/li>\n<li><strong>The coil rule.<\/strong> A current running round a coil makes one end of it a north pole.<\/li>\n<\/ul>\n\n<p>All three are the same cross product wearing different clothes, and your hand is simply how you read a cross product off a flat page. Learn one properly and the other two cost you nothing.<\/p>\n\n<h2>The Right-Hand Rule Formula and What Every Symbol Means<\/h2>\n\n<p>The force rule is one equation, and it is a cross product rather than an ordinary multiplication.<\/p>\n\n<div class=\"pf-formula\">F = q v \u00d7 B<\/div>\n\n<p>The cross is doing real work there. Written out one component at a time, the product of the velocity and the field is three separate subtractions, and that expansion is what a direction answer is actually made of.<\/p>\n\n<div class=\"pf-formula\">(v \u00d7 B)<sub>x<\/sub> = v<sub>y<\/sub>B<sub>z<\/sub> &#8211; v<sub>z<\/sub>B<sub>y<\/sub>, (v \u00d7 B)<sub>y<\/sub> = v<sub>z<\/sub>B<sub>x<\/sub> &#8211; v<sub>x<\/sub>B<sub>z<\/sub>, (v \u00d7 B)<sub>z<\/sub> = v<sub>x<\/sub>B<sub>y<\/sub> &#8211; v<sub>y<\/sub>B<sub>x<\/sub><\/div>\n\n<p>Doing those three subtractions by hand is where marks get lost, so if you only want the answer you can hand both vectors to our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/right-hand-rule-physics\">Right-Hand Rule Calculator<\/a>, which returns the force vector, its size and its direction in words. The other two rules have their own one-line formulae for the size of the field.<\/p>\n\n<div class=\"pf-formula\">B = \u03bc<sub>0<\/sub> I \/ 2\u03c0r<\/div>\n\n<div class=\"pf-formula\">B = \u03bc<sub>0<\/sub> n I<\/div>\n\n<p>And the motor rule, the force on a current-carrying wire, is the force rule again with the charge-times-velocity replaced by the current times the length of wire in the field.<\/p>\n\n<div class=\"pf-formula\">F = I L \u00d7 B<\/div>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;word-break:break-word;\">\n<thead>\n<tr><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">Symbol<\/th><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">Meaning<\/th><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">SI unit<\/th><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">Values used on this page<\/th><\/tr>\n<\/thead>\n<tbody>\n<tr><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">F<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">force on the charge or on the wire, a vector<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">newton, N<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">1.923 \u00d7 10<sup>-13<\/sup> N on a proton; 0.450 N on a wire; 0.200 N in problem 7; exactly 0 N wherever v lies along B<\/td><\/tr>\n<tr><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">q<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">charge, with its sign kept<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">coulomb, C<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">+1.602176634 \u00d7 10<sup>-19<\/sup> C for a proton, minus that for an electron, and 2.0 \u00d7 10<sup>-6<\/sup> C in problem 7<\/td><\/tr>\n<tr><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">v<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">velocity of the charge, a vector<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">metre per second, m\/s<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">3.0 \u00d7 10<sup>6<\/sup> m\/s, and the off-axis (3.0 \u00d7 10<sup>5<\/sup>, 4.0 \u00d7 10<sup>5<\/sup>, 0) m\/s in problem 7<\/td><\/tr>\n<tr><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">B<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">magnetic flux density, a vector<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">tesla, T<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">0.40 T, 0.60 T, 0.20 T, 40.0 \u00b5T, 2.5133 mT, 5.0265 mT<\/td><\/tr>\n<tr><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">I<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">conventional current, positive the way positive charge would flow<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">ampere, A<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">10 A in a wire, 2.0 A in a coil, 3.0 A in a motor wire<\/td><\/tr>\n<tr><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">L<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">length of wire inside the field, pointing the way the current runs<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">metre, m<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">0.25 m<\/td><\/tr>\n<tr><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">r<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">distance from the wire to the point you are standing at<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">metre, m<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">0.050 m<\/td><\/tr>\n<tr><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">n<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">turns per metre along the coil<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">per metre, \/m<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">1000 \/m and 2000 \/m<\/td><\/tr>\n<tr><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">\u03bc<sub>0<\/sub><\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">magnetic constant, the permeability of free space<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">tesla metre per ampere, T\u00b7m\/A<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">4\u03c0 \u00d7 10<sup>-7<\/sup> T\u00b7m\/A<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p>One caution about that last row. Since the 2019 revision of the SI, \u03bc<sub>0<\/sub> is a measured quantity rather than a defined one, and 4\u03c0 \u00d7 10<sup>-7<\/sup> T\u00b7m\/A agrees with the measured value to within about a part in 10<sup>9<\/sup>. The bound is deliberately loose, because the measured figure has shifted between data releases and no value was sourced for this page; use 4\u03c0 \u00d7 10<sup>-7<\/sup>, but do not call it exact.<\/p>\n\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Right-Hand Rule Lab<\/span><\/div><div class=\"pf-sim-slot-body\"><style>.pf-sim-frame{width:100%;border:none;height:600px}@media(max-width:760px){.pf-sim-frame{height:1000px}}<\/style><iframe src=\"\/labs\/right-hand-rule-physics.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n\n<h2>How to Point Your Hand: the Three Rules in Order<\/h2>\n\n<p>Hold your right hand flat, fingers together, thumb stuck out sideways at ninety degrees to them: that one shape, turned to face the problem, solves all three rules.<\/p>\n\n<figure style=\"margin:32px auto;max-width:720px;text-align:center;\">\n  <img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/10\/right-hand-rule-physics-diagram-01-768x564.jpg\" srcset=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/10\/right-hand-rule-physics-diagram-01-768x564.jpg 768w, https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/10\/right-hand-rule-physics-diagram-01.jpg 1400w\" sizes=\"auto, (max-width: 760px) 100vw, 720px\" width=\"1400\" height=\"1028\" alt=\"Three panels on a cream background, each using the same schematic right hand a different way. Panel one, headed force on a moving charge, carries the formula F equals q v cross B: the hand has its palm at the left, four finger bars pointing right and a thumb bar pointing down, with a dashed arc curving away from the fingertips towards the reader labelled curl towards B, out of the page. Beneath it a vector triad leaves one dot: a navy arrow pointing right labelled v right plus x, a thick black arrow pointing down labelled F down minus y, and three small circles each with a filled centre dot standing for a field coming out of the page. Panel two, headed field around a straight wire, carries B equals mu nought I over two pi r: the same hand with its fingers pointing down and its thumb pointing right along the current, and below it a horizontal wire with a solid arrowhead at its right end labelled I, a circled dot above the wire labelled B out of the page if you stand above, and a circled cross below the wire labelled B into the page if you stand below. Panel three, headed which end of a coil is north, carries B equals mu nought n I: the same hand with fingers down and thumb right, and below it a solenoid drawn as six turns, each turn a solid front strand carrying a downward arrowhead and a dashed back strand, with S at the left end, N at the right end and two wine arrows through the bore pointing right under the caption B inside runs S to N, so it points right. Three footnotes state that arrow lengths are indicative only and nothing is drawn to scale, that panel one is a positive charge and reversing the charge reverses F while the hand never changes, and that current downwards on the front of each turn is anticlockwise seen from the right-hand end, which puts north on the right.\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;border-radius:4px;\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">The same right hand, used three ways. Nothing in the figure is drawn to scale, and the arrow lengths carry no information; what they carry is direction, and every direction in all three panels obeys the axis convention printed at the top.<\/figcaption>\n<\/figure>\n\n<h3>Rule 1: the force on a moving charge<\/h3>\n\n<p>Point the fingers of your right hand the way the charge is moving, curl them towards the field, and your thumb now lies along the force on a <em>positive<\/em> charge. Keep the hand exactly as it is for a negative charge and simply reverse the answer at the end.<\/p>\n\n<p>Work the standard case. A proton moving to your right through a field coming out of the page: fingers right, curl them towards you, and the thumb drops. The force is downwards, and nothing you do with your left hand will make it any more downwards.<\/p>\n\n<h3>Rule 2: the field around a straight wire<\/h3>\n\n<p>Grip the wire with your right hand so that your thumb lies along the conventional current, and your curled fingers now run the way the field circles the wire. Read the field at any point by seeing which way your fingers are heading as they pass it.<\/p>\n\n<p>Conventional current is the one that matters here, and it runs the opposite way to the drift of the electrons that are really moving. Get that backwards and the field circles the wrong way; see our guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/electric-current\/\">electric current<\/a> for why the convention is the way round it is.<\/p>\n\n<h3>Rule 3: which end of a coil is north<\/h3>\n\n<p>Wrap your right hand round the coil with your fingers curling the way the current runs round the turns, and your thumb sticks out of the north end. That is all there is to a <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/solenoid\/\">solenoid<\/a>: a bar magnet you can switch off, with its poles set by which way you wound it.<\/p>\n\n<p>The size of the field inside a long coil needs no hand at all, only the turns per metre and the current, and our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/magnetic-field\">Magnetic Field Calculator<\/a> will do that arithmetic for you.<\/p>\n\n<h3>The three rules side by side<\/h3>\n\n<p>All three rules differ only in which part of the hand you set and which part you read back off, which is why one grid holds them all. Notice that the input and the output swap places between the second rule and the third: the thumb is what you set for a wire, and what you read for a coil.<\/p>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;word-break:break-word;\">\n<thead>\n<tr><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">Rule<\/th><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">The thumb<\/th><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">The fingers<\/th><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">Formula<\/th><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">What you read off<\/th><\/tr>\n<\/thead>\n<tbody>\n<tr><th style=\"text-align:left;padding:8px 10px;border-bottom:1px solid #D9CFB8;font-weight:700;\">Force on a moving charge<\/th><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">Lies along the force when the charge is positive<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">Set along the velocity, then curled towards the field<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">F = q v \u00d7 B<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">The thumb: which way the charge is pushed. Turn the answer round if q is negative.<\/td><\/tr>\n<tr><th style=\"text-align:left;padding:8px 10px;border-bottom:1px solid #D9CFB8;font-weight:700;\">Field around a straight wire<\/th><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">Set along the conventional current in the wire<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">Wrapped round the wire, so they trace the field circles<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">B = \u03bc<sub>0<\/sub> I \/ 2\u03c0r<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">The fingers: which way the field heads as it passes the point you chose.<\/td><\/tr>\n<tr><th style=\"text-align:left;padding:8px 10px;border-bottom:1px solid #D9CFB8;font-weight:700;\">North end of a coil<\/th><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">Sticks out of whichever end is north<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">Curled in the sense the current circulates round the windings<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">B = \u03bc<sub>0<\/sub> n I<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">The thumb: which end of the coil is north, and so which end is south.<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p>The formula column is there to show that the size is a separate question with a separate answer. No hand position produces a number, and nothing in those three formulae tells you a direction.<\/p>\n\n<h3>Three ways to hold the same hand<\/h3>\n\n<p>You will meet the force rule in at least three forms, and students sometimes think they contradict each other. They do not: they are three ways of pointing the same hand.<\/p>\n\n<ul>\n<li><strong>Fingers then curl.<\/strong> Fingers along v, curl them towards B, thumb along F. The form used throughout this page.<\/li>\n<li><strong>Flat hand.<\/strong> Fingers along v, palm facing the way B points, thumb along F. Quicker once the field is coming straight at you.<\/li>\n<li><strong>Three fingers.<\/strong> Thumb along v, first finger along B, second finger along F, all three held at right angles. That is the same right-handed set as the x, y and z axes themselves, which is where the hand came from in the first place.<\/li>\n<\/ul>\n\n<p>Fleming&#8217;s left-hand rule belongs in the same list, and it is the one place the hand genuinely changes. It assigns the thumb to the force, the first finger to the field and the second finger to the current \u2014 the three fingers above with two of them swapped over \u2014 and swapping two members of a right-handed set makes it left-handed, so it needs the other hand to come out right. A cyclic shuffle would have left it right-handed; this one is not cyclic.<\/p>\n\n<p>The answers are identical either way. Fleming&#8217;s rule is a different mnemonic for the motor case, not a different physics, and nothing on this page contradicts it.<\/p>\n\n<h2>The Six Directions, and the 36 Pairs They Make<\/h2>\n\n<figure style=\"margin:32px auto;max-width:720px;text-align:center;\">\n  <img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/10\/right-hand-rule-physics-diagram-02-768x474.jpg\" srcset=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/10\/right-hand-rule-physics-diagram-02-768x474.jpg 768w, https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/10\/right-hand-rule-physics-diagram-02.jpg 1400w\" sizes=\"auto, (max-width: 760px) 100vw, 720px\" width=\"1400\" height=\"864\" alt=\"A two-panel reference figure on a cream background. The left panel, headed the six directions all from one origin, has four navy arrows leaving a single dot: one to the right labelled right plus x, one to the left labelled left minus x, one upwards labelled up plus y and one downwards labelled down minus y. Up and to the left of the origin a circle with a filled centre dot is labelled out of the page plus z; down and to the right a circle with a diagonal cross through it is labelled into the page minus z. A line underneath says that z points at you, so it is drawn as a symbol and never as an arrow in the plane. The right panel, headed right-handed so the order matters, lists four cross products in bold with the direction spelled out beneath each: x cross y equals plus z, out of the page; y cross z equals plus x, right; z cross x equals plus y, up; and, below a rule and marked swap the order and it reverses, y cross x equals minus z, into the page. Below a second rule, x cross x equals zero, with the note that a vector crossed with itself is zero and zero has no direction. Three footnotes state that 12 of the 36 ordered pairs of these six directions give a zero cross product, three axes by two senses of v by two senses of B; that every direction named in the article means exactly the arrow or symbol drawn here; and that the products above are geometry and do not depend on the charge, while it is the force q v cross B that reverses when q is negative.\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;border-radius:4px;\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">The six directions this page ever names, and the three products that fix the sense of all the others. The z axis points at the reader, so it is drawn as a circled dot coming out and a circled cross going in, never as an arrow lying in the plane.<\/figcaption>\n<\/figure>\n\n<p>Restrict the velocity and the field to those six directions and there are 36 ordered pairs to consider. The table below gives the force direction for every one of them, for a positive charge.<\/p>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;word-break:break-word;\">\n<thead>\n<tr><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">v points<\/th><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">B right (+x)<\/th><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">B left (-x)<\/th><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">B up (+y)<\/th><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">B down (-y)<\/th><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">B out of the page (+z)<\/th><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">B into the page (-z)<\/th><\/tr>\n<\/thead>\n<tbody>\n<tr><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">right (+x)<\/th><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;background:#F5F2EA;font-style:italic;\">no force<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;background:#F5F2EA;font-style:italic;\">no force<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">out of the page (+z)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">into the page (-z)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">down (-y)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">up (+y)<\/td><\/tr>\n<tr><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">left (-x)<\/th><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;background:#F5F2EA;font-style:italic;\">no force<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;background:#F5F2EA;font-style:italic;\">no force<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">into the page (-z)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">out of the page (+z)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">up (+y)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">down (-y)<\/td><\/tr>\n<tr><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">up (+y)<\/th><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">into the page (-z)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">out of the page (+z)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;background:#F5F2EA;font-style:italic;\">no force<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;background:#F5F2EA;font-style:italic;\">no force<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">right (+x)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">left (-x)<\/td><\/tr>\n<tr><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">down (-y)<\/th><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">out of the page (+z)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">into the page (-z)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;background:#F5F2EA;font-style:italic;\">no force<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;background:#F5F2EA;font-style:italic;\">no force<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">left (-x)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">right (+x)<\/td><\/tr>\n<tr><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">out of the page (+z)<\/th><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">up (+y)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">down (-y)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">left (-x)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">right (+x)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;background:#F5F2EA;font-style:italic;\">no force<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;background:#F5F2EA;font-style:italic;\">no force<\/td><\/tr>\n<tr><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;\">into the page (-z)<\/th><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">down (-y)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">up (+y)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">right (+x)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">left (-x)<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;background:#F5F2EA;font-style:italic;\">no force<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;background:#F5F2EA;font-style:italic;\">no force<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p>Two things in that table are worth more than the individual cells. The first is that <strong>12 of the 36 pairs give no force at all<\/strong> \u2014 three axes, two senses of the velocity and two senses of the field. It is a third of the table, not a sixth, and the count catches people out because it is easy to forget that each axis can be travelled along in two directions.<\/p>\n\n<p>The second is the symmetry. Every cell is reversed by reversing the sign of the charge, so one table serves for electrons too: read the cell, then turn the answer round. The simulator above is restricted to these same six directions on purpose, which keeps it a direction trainer; anything off-axis belongs in the calculator, and problem 7 below works one through by hand.<\/p>\n\n<h2>Why a Cross Product Needs a Hand<\/h2>\n\n<p>Two vectors can be combined in two quite different ways, and the difference is the whole of this topic. A dot product returns a number and does not care about the order. A cross product returns a vector perpendicular to both, and the order decides which of the two perpendicular ways it points.<\/p>\n\n<p>That last sentence is the problem in miniature. Take any two vectors that are not parallel; they define a plane, and the plane has two faces. Nothing in the arithmetic of components prefers one face over the other \u2014 the preference has to be put in by hand, and it is put in by the right hand.<\/p>\n\n<p>Done consistently in a left-handed coordinate system the left hand would serve just as well, and every answer would come out the same. The hand is bookkeeping about the axes, not a fact about hands, which is why it appears identically in <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/torque-physics\/\">torque<\/a> and in angular velocity, neither of which involves magnetism at all.<\/p>\n\n<h3>Why the force never does any work<\/h3>\n\n<p>Because F is a cross product of v with something, it can never have a component along v. Over all 36 axis pairs the dot product F \u00b7 v is exactly zero, with no rounding involved. The dot product is the honest way to put it, because in the 12 pairs where the force vanishes there is no angle to take the cosine of.<\/p>\n\n<p>That has a consequence worth holding on to: a magnetic field can bend a charge&#8217;s path but can never change its speed. It is the same point our <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/magnetic-field\/\">magnetic field<\/a> guide makes when it says a field steers but never accelerates in the everyday sense.<\/p>\n\n<p>A charge released into a steady uniform field therefore travels at constant speed round a circle, and a magnetic field can sort particles by bending them while leaving their kinetic energy untouched.<\/p>\n\n<h3>Crossing is not adding<\/h3>\n\n<p>One last distinction, because it trips people who are new to vectors. Adding two vectors gives you something that lies between them; crossing them gives you something that lies outside their plane altogether, perpendicular to both.<\/p>\n\n<p>If what your problem actually wants is a resultant rather than a perpendicular, you need a different tool and a different idea \u2014 our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/vector-addition\">Vector Addition Calculator<\/a> handles that case, and the difference between the two is covered in our guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/kinematics\/scalar-vector-quantities\/\">scalar and vector quantities<\/a>.<\/p>\n\n<h2>Common Mistakes With the Right-Hand Rule<\/h2>\n\n<p>Four mistakes account for most wrong answers, and three of them are about signs rather than about the hand.<\/p>\n\n<h3>Mistake 1: guessing the sense of z<\/h3>\n\n<p>A field drawn as a grid of crosses is going <em>into<\/em> the page and must be entered as a negative z component; a grid of dots is coming out at you and is positive. Swap those two and every direction you work out comes back reversed, which is why this page fixes the convention before it names a single direction.<\/p>\n\n<h3>Mistake 2: switching to the left hand for an electron<\/h3>\n\n<p>Changing hands for a negative charge happens to give the right answer, and that is exactly what makes it dangerous: it works until it is combined with another sign change, and then it quietly fails. Use the right hand always, then reverse the result once because q is negative.<\/p>\n\n<h3>Mistake 3: using electron flow instead of conventional current<\/h3>\n\n<p>The grip rule and the coil rule are both stated for conventional current, which flows from the positive terminal round to the negative. Electrons drift the other way. Trace the electrons with your thumb and you will get the field circling backwards and the coil&#8217;s poles swapped.<\/p>\n\n<h3>Mistake 4: thinking six of the 36 pairs give zero<\/h3>\n\n<p>Three axes suggests three zero cases, and remembering that parallel and antiparallel both count suggests six. The real figure is 12, because the velocity has two senses and the field has two senses, and 3 \u00d7 2 \u00d7 2 = 12. Count the italic cells in the table above if you want to check.<\/p>\n\n<h2>Where the Right-Hand Rule Breaks Down<\/h2>\n\n<p>The hand itself never breaks down \u2014 a cross product is a cross product. What has limits is the physics the three formulae describe, and it is worth knowing where those limits sit.<\/p>\n\n<h3>A zero force has no direction, and that is an answer<\/h3>\n\n<p>When the charge, the velocity or the field is zero, or when the velocity lies exactly along the field, the cross product is the zero vector. There is no axis to name and no angle to quote, so the honest output is the cause rather than an invented direction. Any tool that prints a direction here is telling you something it cannot know.<\/p>\n\n<h3>The straight-wire formula assumes an infinitely long wire<\/h3>\n\n<p>B = \u03bc<sub>0<\/sub> I \/ 2\u03c0r is derived for a wire of infinite length, and a real wire departs from it near its ends. The direction the grip rule gives is unaffected; it is the size that drifts, and only close to where the wire stops.<\/p>\n\n<h3>The solenoid formula is the ideal long-coil interior<\/h3>\n\n<p>B = \u03bc<sub>0<\/sub> n I is the field well inside a long, tightly wound coil. Near the ends it falls away, and for a short fat coil it is wrong everywhere. Which end is north does not depend on any of that \u2014 only the number does.<\/p>\n\n<h3>One charge, one uniform field, no relativity<\/h3>\n\n<p>Every figure on this page treats a single charge in a field that is the same everywhere, moving slowly enough that no relativistic correction is needed. At 3.0 \u00d7 10<sup>6<\/sup> m\/s a charge is at about one per cent of the speed of light, where that is comfortably true.<\/p>\n\n<p>There is also no electric field anywhere in these problems. The complete force on a charge is the Lorentz force, of which the magnetic half is what the hand gives you; if your problem has an electric field as well, its contribution has to be added separately, and unlike the magnetic part it does do work.<\/p>\n\n<h2>Where the Right-Hand Rule Is Actually Used<\/h2>\n\n<p>This is not a rule you learn and then file away. It decides which way real things move, and getting it backwards is the difference between a motor that spins and a motor that spins the wrong way.<\/p>\n\n<h3>Motors, and the direction a shaft turns<\/h3>\n\n<p>A motor is a loop of wire carrying current across a magnetic field, and the force on each side of the loop comes straight from F = I L \u00d7 B. The two sides carry current in opposite directions, so the forces on them point opposite ways, and the loop turns. Which way it turns is the right-hand rule, applied twice.<\/p>\n\n<figure style=\"margin:32px auto;max-width:640px;text-align:center;\">\n  <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/10\/a-small-DC-motor-opened-up.avif\"\n       alt=\"A small brushed DC motor opened up, showing the wire coil of the rotor between two curved permanent magnets.\"\n       loading=\"lazy\"\n       style=\"width:100%;height:auto;border-radius:4px;\" \/>\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">Every brushed DC motor is the right-hand rule made of metal: current through the coil, a field across it, and a force at right angles to both.<\/figcaption>\n<\/figure>\n\n<h3>Reading the sign of the carriers in a Hall probe<\/h3>\n\n<p>Push a current through a thin strip in a transverse field and the moving carriers pile up on one edge. Which edge depends on the sign of the carriers, so measuring the voltage across the strip tells you whether the charge doing the moving is positive or negative. That is a direction question answered with a direction rule.<\/p>\n\n<h3>Electromagnets, relays and the pole you need<\/h3>\n\n<p>Any coil you wind becomes a magnet with its poles decided by the direction you wound it, and relays, solenoid valves and lifting magnets all depend on getting that right. Reverse the current and the poles swap, instantly.<\/p>\n\n<h3>Induction, and which way the induced current runs<\/h3>\n\n<p>When a field through a loop changes, the loop pushes back: the induced current runs the way that opposes the change, and the right-hand rule is how you turn &#8220;opposes&#8221; into an actual direction round the loop. Our guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/electromagnetic-induction\/\">electromagnetic induction<\/a> works that argument through. If you want to go further with the trainer above, the <a href=\"https:\/\/physicsfundamentalsinfo.com\/labs\/right-hand-rule-physics-simulator\">right-hand rule simulator page<\/a> covers what each control changes and which preset to load first.<\/p>\n\n<h2>Worked Problems<\/h2>\n\n<p>Every problem below asks for a direction first and a size second, and every figure is quoted from the same physics contract the simulator and the calculator are both built against. Axes as before: +x right, +y up, +z out of the page.<\/p>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">A proton travels to the right (+x) at 3.0 x 10^6 m\/s through a uniform 0.40 T field pointing out of the page (+z). Which way does the magnetic force on it point, and how large is that force?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: The force is the cross product F = q(v \u00d7 B), so start from the component expansion. With v along +x only and B along +z only, two of the three components vanish and only the y component survives: (v \u00d7 B)<sub>y<\/sub> = v<sub>z<\/sub>B<sub>x<\/sub> &#8211; v<sub>x<\/sub>B<sub>z<\/sub>.\n\nStep 2: Substitute, in SI units. (v \u00d7 B)<sub>y<\/sub> = (0)(0) &#8211; (3.0 \u00d7 10<sup>6<\/sup>)(0.40) = -1.2 \u00d7 10<sup>6<\/sup> T\u00b7m\/s.\n\nStep 3: Multiply by the charge of a proton, q = +1.602176634 \u00d7 10<sup>-19<\/sup> C. F<sub>y<\/sub> = (1.602176634 \u00d7 10<sup>-19<\/sup>)(-1.2 \u00d7 10<sup>6<\/sup>) = -1.92261 \u00d7 10<sup>-13<\/sup> N, with F<sub>x<\/sub> and F<sub>z<\/sub> both zero. A negative y component means the force points down.\n\n<strong>Answer: 1.923 \u00d7 10<sup>-13<\/sup> N, pointing down (-y).<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">Replace the proton in problem 1 with an electron, keeping the same velocity and the same field. What changes?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Nothing about the hand changes, and nothing about the cross product changes either. The velocity and the field are the same, so (v \u00d7 B)<sub>y<\/sub> = -1.2 \u00d7 10<sup>6<\/sup> T\u00b7m\/s exactly as before.\n\nStep 2: Only the charge is different: q = -1.602176634 \u00d7 10<sup>-19<\/sup> C.\n\nStep 3: F<sub>y<\/sub> = (-1.602176634 \u00d7 10<sup>-19<\/sup>)(-1.2 \u00d7 10<sup>6<\/sup>) = +1.92261 \u00d7 10<sup>-13<\/sup> N. Two negatives have made a positive, so the force now points up.\n\n<strong>Answer: 1.923 \u00d7 10<sup>-13<\/sup> N, pointing up (+y) \u2014 the same size as for the proton, the opposite way.<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">A proton again travels to the right (+x) at 3.0 x 10^6 m\/s, but this time the 0.40 T field points into the page (-z). Which way is the force now?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Into the page means a negative z component, so B<sub>z<\/sub> = -0.40 T. The surviving component of the cross product is the same one: (v \u00d7 B)<sub>y<\/sub> = v<sub>z<\/sub>B<sub>x<\/sub> &#8211; v<sub>x<\/sub>B<sub>z<\/sub>.\n\nStep 2: (v \u00d7 B)<sub>y<\/sub> = (0)(0) &#8211; (3.0 \u00d7 10<sup>6<\/sup>)(-0.40) = +1.2 \u00d7 10<sup>6<\/sup> T\u00b7m\/s.\n\nStep 3: F<sub>y<\/sub> = (1.602176634 \u00d7 10<sup>-19<\/sup>)(+1.2 \u00d7 10<sup>6<\/sup>) = +1.92261 \u00d7 10<sup>-13<\/sup> N. Reversing the field has reversed the force, and the size has not moved.\n\n<strong>Answer: 1.923 \u00d7 10<sup>-13<\/sup> N, pointing up (+y). Half of this answer was the sense of z.<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">A long straight wire carries 10 A to the right (+x). Which way does its magnetic field point at a point 0.050 m directly above the wire, and at a point 0.050 m directly below it, and how strong is the field at each?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Direction first, from the grip rule written as a cross product: B points along the current direction crossed with the direction from the wire to you. Standing above the wire that is (+x) crossed with (+y), which is +z, out of the page. Standing below it is (+x) crossed with (-y), which is -z, into the page.\n\nStep 2: Size next, from B = \u03bc<sub>0<\/sub> I \/ 2\u03c0r. Because \u03bc<sub>0<\/sub> = 4\u03c0 \u00d7 10<sup>-7<\/sup> T\u00b7m\/A, the combination \u03bc<sub>0<\/sub> \/ 2\u03c0 is 2 \u00d7 10<sup>-7<\/sup> T\u00b7m\/A, so B = (2 \u00d7 10<sup>-7<\/sup>)(10) \/ (0.050).\n\nStep 3: (2 \u00d7 10<sup>-7<\/sup>)(10) = 2 \u00d7 10<sup>-6<\/sup>, and 2 \u00d7 10<sup>-6<\/sup> \/ 0.050 = 4.0 \u00d7 10<sup>-5<\/sup> T. Both points are the same distance from the wire, so both get the same size.\n\n<strong>Answer: out of the page (+z) above the wire, into the page (-z) below it, and 40.0 \u00b5T at both points.<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">A straight wire 0.25 m long carries 3.0 A to the right (+x) through a 0.60 T field pointing into the page (-z). Which way is the force on the wire, and how big is it?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: For a current-carrying wire the rule is F = I L \u00d7 B, which is the force rule with the charge-times-velocity replaced by current-times-length. Same hand, same expansion, so the surviving component is (L \u00d7 B)<sub>y<\/sub> = L<sub>z<\/sub>B<sub>x<\/sub> &#8211; L<sub>x<\/sub>B<sub>z<\/sub>.\n\nStep 2: With L<sub>x<\/sub> = 0.25 m and B<sub>z<\/sub> = -0.60 T, (L \u00d7 B)<sub>y<\/sub> = (0)(0) &#8211; (0.25)(-0.60) = +0.15 T\u00b7m.\n\nStep 3: F<sub>y<\/sub> = I(L \u00d7 B)<sub>y<\/sub> = (3.0)(0.15) = 0.450 N, upwards. Cross-check against the textbook size for a wire at right angles to the field, BIL: (0.60)(3.0)(0.25) = 0.450 N, the same figure.\n\n<strong>Answer: 0.450 N, straight up (+y).<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">A long solenoid has 1000 turns per metre and carries 2.0 A, running anticlockwise as seen from its right-hand end. Which end is the north pole, and what is the field inside? What happens if the winding density is doubled to 2000 turns per metre?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Direction first. Curl the fingers of your right hand the way the conventional current runs round the turns and your thumb comes out of the north end. Anticlockwise as seen from the right-hand end sends the thumb towards +x, so north is the right-hand end and south is the left.\n\nStep 2: Size, from B = \u03bc<sub>0<\/sub> n I. B = (4\u03c0 \u00d7 10<sup>-7<\/sup>)(1000)(2.0) = 2.51327 \u00d7 10<sup>-3<\/sup> T.\n\nStep 3: Double the turns per metre and nothing about the geometry moves, so only the number changes: B = (4\u03c0 \u00d7 10<sup>-7<\/sup>)(2000)(2.0) = 5.026548 \u00d7 10<sup>-3<\/sup> T. The field has doubled; the north end has not moved.\n\n<strong>Answer: north at the right-hand end, with 2.5133 mT inside at 1000 turns per metre and 5.0265 mT at 2000.<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A charge of 2.0 x 10^-6 C moves with velocity (3.0 x 10^5, 4.0 x 10^5, 0) m\/s through a field (0, 0, 0.20) T. Find the force vector, its size, and show that the force does no work.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Neither vector lies along an axis, so all three components of the cross product have to be worked out. (v \u00d7 B)<sub>x<\/sub> = (4.0 \u00d7 10<sup>5<\/sup>)(0.20) &#8211; (0)(0) = 8.0 \u00d7 10<sup>4<\/sup> T\u00b7m\/s; (v \u00d7 B)<sub>y<\/sub> = (0)(0) &#8211; (3.0 \u00d7 10<sup>5<\/sup>)(0.20) = -6.0 \u00d7 10<sup>4<\/sup> T\u00b7m\/s; (v \u00d7 B)<sub>z<\/sub> = (3.0 \u00d7 10<sup>5<\/sup>)(0) &#8211; (4.0 \u00d7 10<sup>5<\/sup>)(0) = 0.\n\nStep 2: Multiply all three by q = 2.0 \u00d7 10<sup>-6<\/sup> C. F<sub>x<\/sub> = (2.0 \u00d7 10<sup>-6<\/sup>)(8.0 \u00d7 10<sup>4<\/sup>) = 0.160 N; F<sub>y<\/sub> = (2.0 \u00d7 10<sup>-6<\/sup>)(-6.0 \u00d7 10<sup>4<\/sup>) = -0.120 N; F<sub>z<\/sub> = 0.\n\nStep 3: The size is sqrt(F<sub>x<\/sub><sup>2<\/sup> + F<sub>y<\/sub><sup>2<\/sup> + F<sub>z<\/sub><sup>2<\/sup>) = sqrt(0.0256 + 0.0144 + 0) = sqrt(0.0400) = 0.200 N. Dividing the components by that size gives the unit vector (0.800, -0.600, 0.000), which is none of the six named directions. For the no-work check, multiply the force by the velocity component by component: (0.160)(3.0 \u00d7 10<sup>5<\/sup>) + (-0.120)(4.0 \u00d7 10<sup>5<\/sup>) + 0 = 48000 &#8211; 48000 = 0.\n\n<strong>Answer: F = (0.160, -0.120, 0) N, size 0.200 N, direction the unit vector (0.800, -0.600, 0.000), and the force does no work.<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 8<\/div><div class=\"pf-problem-question\">A proton travels to the right (+x) at 3.0 x 10^6 m\/s. What is the magnetic force on it if the 0.40 T field also points to the right, and what if the field points to the left instead?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Field to the right, so B = (0.40, 0, 0) T. Every component of the cross product has a zero in each of its terms: (v \u00d7 B)<sub>x<\/sub> = (0)(0) &#8211; (0)(0) = 0; (v \u00d7 B)<sub>y<\/sub> = (0)(0.40) &#8211; (3.0 \u00d7 10<sup>6<\/sup>)(0) = 0; (v \u00d7 B)<sub>z<\/sub> = (3.0 \u00d7 10<sup>6<\/sup>)(0) &#8211; (0)(0.40) = 0.\n\nStep 2: Field to the left, so B = (-0.40, 0, 0) T. The same three lines give (v \u00d7 B)<sub>x<\/sub> = (0)(0) &#8211; (0)(0) = 0, then (v \u00d7 B)<sub>y<\/sub> = (0)(-0.40) &#8211; (3.0 \u00d7 10<sup>6<\/sup>)(0) = 0 and (v \u00d7 B)<sub>z<\/sub> = (3.0 \u00d7 10<sup>6<\/sup>)(0) &#8211; (0)(-0.40) = 0, so the cross product is the zero vector again.\n\nStep 3: A zero vector has no direction, so the complete answer names the cause instead of an axis: in the first case the velocity is parallel to the field, at an angle of 0 degrees, and in the second it is antiparallel, at 180 degrees. This is also why the tools on this site take the force from the cross product rather than from the sine form: in double-precision arithmetic sin(180 degrees) evaluates to 1.2246 \u00d7 10<sup>-16<\/sup> instead of zero, which turns an exact zero into 2.355 \u00d7 10<sup>-29<\/sup> N of floating-point dust.\n\n<strong>Answer: zero force in both cases, and no direction in either.<\/strong>\n\n<\/div><\/details><\/div>\n\n<h2>Frequently Asked Questions<\/h2>\n\n<details class=\"pf-faq-item\"><summary>What is the right-hand rule in physics?<\/summary><div class=\"pf-faq-item-answer\">\n\nThe right-hand rule is a convention for reading the direction of a cross product off your hand. Point your fingers along the first vector, curl them towards the second, and your thumb gives the third. It fixes the direction of the magnetic force on a moving charge, of the field circling a current-carrying wire, and of the north pole of a coil.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Which finger is which in the right-hand rule?<\/summary><div class=\"pf-faq-item-answer\">\n\nIn the form used here the fingers go along the velocity, they curl towards the magnetic field, and the thumb gives the force on a positive charge. A three-finger version of the same rule puts the thumb along the velocity, the first finger along the field and the second finger along the force. Fleming&#8217;s left-hand rule reshuffles those three, which is exactly why it needs the other hand.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Why the right hand and not the left?<\/summary><div class=\"pf-faq-item-answer\">\n\nBecause the standard coordinate axes are right-handed, and the hand is simply bookkeeping about that choice. A cross product has to pick one of the two sides of a plane, and nothing in the arithmetic picks for you. In a consistently left-handed system the left hand would work and every answer would agree. Fleming&#8217;s left-hand rule is the same physics with the fingers assigned differently.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>How do you use the right-hand rule for an electron?<\/summary><div class=\"pf-faq-item-answer\">\n\nUse the right hand exactly as you would for a positive charge, then reverse the answer at the end. The charge sits outside the cross product in F = q v \u00d7 B, so a negative q simply flips the vector the hand produced. Swapping to your left hand happens to work too, but it fails as soon as a second sign change is involved.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Does the right-hand rule work for a current-carrying wire as well as a moving charge?<\/summary><div class=\"pf-faq-item-answer\">\n\nYes, and it is the same rule. Replace charge-times-velocity by current-times-length and the force on a wire is F = I L \u00d7 B, read with the same hand and the same geometry. The grip rule for the field around a wire and the coil rule for a solenoid are also cross products, which is why one hand position covers all of them.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What happens when the velocity and the magnetic field point the same way?<\/summary><div class=\"pf-faq-item-answer\">\n\nThe force is exactly zero, and a zero force has no direction. The same is true when they point in exactly opposite directions, and when the charge, the speed or the field is zero. Of the 36 pairs you can make from six axis directions, 12 give no force at all: three axes, two senses of the velocity, two senses of the field.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Is there more than one right-hand rule?<\/summary><div class=\"pf-faq-item-answer\">\n\nThere are three common ones, and they are the same piece of physics each time. The force rule gives the push on a moving charge, the grip rule gives the field circling a straight wire, and the coil rule gives the north end of a solenoid. All three read a cross product off the hand; only the vectors you feed in are different.\n\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>The right-hand rule turns a cross product into a direction you can point at: fingers along the first vector, curl them towards the second, thumb along the third. This page fixes the axis convention first, then works through all three rules &#8211; the force on a moving charge, the field circling a wire, and which end of a coil is north &#8211; with a 36-cell direction table, eight worked problems, two original diagrams and an interactive trainer.<\/p>\n","protected":false},"author":1,"featured_media":2143,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"class_list":["post-2140","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-electromagnetism"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/2140","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/comments?post=2140"}],"version-history":[{"count":3,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/2140\/revisions"}],"predecessor-version":[{"id":2159,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/2140\/revisions\/2159"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/2143"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=2140"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=2140"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=2140"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}