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14px"><i class="fa fa-fw"></i><a href="../1136/capacitors.html">Capacitors</a></label></li></ul></li></ul></div> <div class="hidden-xs hidden-sm"> <button class="btn btn-default btn-block text-xs-center" data-toggle="modal" data-target="#modal-feedback" style="margin-bottom: 10px"><i class="fa fa-send"></i> Feedback</button> </div> </div> <div class="col-md-9" id="main-column"> <h1 class="page_title"> Lenz's law <a href="#" class="mark-page-favorite pull-right" data-pid="1142" title="Mark as favorite" onclick="return false;"><i class="fa fa-star-o"></i></a> </h1> <ol class="breadcrumb"> <li><a href="../../../physics.html"><i class="fa fa-home"></i></a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../445/electricity-and-magnetism.html">Electricity and magnetism</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../1144/ahl-em-induction.html">AHL EM induction</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">Lenz's law</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 20 minutes"><i class="fa fa-clock-o"></i> 20'</span> </ol> <article id="main-article"> <p><img alt="" src="../../em-induction/lenz.jpg" style="float: left; width: 200px; height: 222px;">Faraday's law gives the magnitude of the EMF induced due to a change in flux. Lenz's law tells us the direction.</p> <hr class="hidden-separator"> <div class="panel panel-turquoise panel-has-colored-body"> <div class="panel-heading"> <div> <p>Key Concepts</p> </div> </div> <div class="panel-body"> <div> <p>Lenz's law states that the direction of the induced current is such that it will oppose the change producing it. This means that we can now state the equation for the induced EMF:</p> <p style="text-align: center;"><span class="math-tex">\(\varepsilon =-{\mathrm{d}N\Phi\over \mathrm{d}t}\)</span></p> <ul> <li><span class="math-tex">\(\varepsilon \)</span> is the induced EMF (V)</li> <li><span class="math-tex">\(N\Phi\)</span> is the magnetic flux linkage (Tm<sup>2</sup> or Wb)</li> <li><span class="math-tex">\({\mathrm{d}N\Phi\over \mathrm{d}t}\)</span> is the rate of change of magnetic flux linkage (Tm<sup>2</sup>s<sup>-1</sup> or Wb s<sup>-1</sup>)</li> </ul> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369633686"></iframe></div> </div> <div class="panel panel-turquoise panel-has-colored-body panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Sliding wire</p> </div> </div> <div class="panel-body"> <div> <p>To fully understand the cause of Lenz's law, we must consider the motion and forces exerted by individual charges.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369633923"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <p> </p> <p> </p> <div class="panel panel-turquoise panel-has-colored-body panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Conservation of energy</p> </div> </div> <div class="panel-body"> <div> <p>Lenz's law is a consequence of <a href="../251/conservation-of-energy.html" title="Conservation of energy">Conservation of energy</a>. An induced EMF does work on charges causing them to move in a net direction (and hence for a current to flow).</p> <p>The induction of current requires energy. Therefore, either:</p> <ul> <li>work must be done to continue the relative motion of the magnetic field and conductor</li> <li>the conductor or magnet will lose kinetic energy so that the EMF induced decreases</li> </ul> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <p> </p> <p> </p> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Essentials</p> </div> </div> <div class="panel-body"> <div> <div class="panel panel-has-colored-body panel-has-border panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Induced EMF direction</p> </div> </div> <div class="panel-body"> <div> <p>Lenz's law has implications for any instance of electromagnetic induction. There are two mechanisms for determining the direction of the EMF:</p> <ol> <li>Consider the direction in which individual moving charges experience a force. This is directed by Fleming's left hand rule.</li> </ol> <p style="text-align: center;"><img alt="" src="../../em-induction/lefthandoutline.png" style="width: 250px; height: 182px;"></p> <ol start="2"> <li>Consider the direction in which EMF is induced by a conductor moving relative to a magnetic field. The switch to the right hand rule incorporates the opposing nature set by Lenz's law.</li> </ol> <p style="text-align: center;"><img alt="" src="../../em-induction/righthandoutline.png" style="width: 250px; height: 182px;"></p> <div class="magenta"> <p>It can sometimes be difficult for learners to work out the EMF direction because of uncertainty in whether to consider individual charges or the whole of the conductor.</p> <p>The good news is that there is a simple solution: since we know the consequence of Lenz's law is to oppose the change producing it, the resulting electromagnet formed will always:</p> <ul> <li>repel an incoming magnetic field</li> <li>attract an outgoing magnetic field</li> </ul> </div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Falling magnet</p> </div> </div> <div class="panel-body"> <div> <p>When a magnet falls through a conductor, the current direction in the conductor is such as to exert an upward force on the magnet.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369633753"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Coil and magnet</p> </div> </div> <div class="panel-body"> <div> <p>When a magnet moves towards a coil, the current direction in the coil is such as to repel the approaching magnet.</p> <p>If a north pole moves to the right towards a coil, the coil will become an electromagnet with a north pole on its left.</p> <p style="text-align: center;"><img alt="" src="../../em-induction/coil-and-magnet.jpg" style="width: 350px; height: 200px;"></p> <p>The right hand grip rule can be used to verify that the direction of conventional current would do so. Note that the current in the side of the coil nearest the reader is upward.</p> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Coil in a changing field</p> </div> </div> <div class="panel-body"> <div> <p>When a coil is placed next to an electromagnet coil, the field produced by the induction in the second coil will be in the opposite direction to the field produced by the first coil.</p> <p style="text-align: center;"><img alt="" class="gifffer" data-gifffer="/media/physics/em-induction/induction1.gif" style="width: 350px; height: 192px;"></p> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Eddy currents</p> </div> </div> <div class="panel-body"> <div> <p>Eddy currents are induced within a conductor. They flow in closed loops that is perpendicular to the magnetic field. These currents themselves cause the conductor to become an electromagnet.</p> <p style="text-align: center;"><img alt="" src="../../em-induction/eddy.svg" style="width: 350px; height: 195px;"></p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369633475"></iframe></div> <p>Eddy currents cause dissipation of thermal energy in devices that employ electromagnetic induction. This can be useful in braking but wasteful in generators and transformers. One mechanism to reduce eddy currents is to laminate any soft iron core present.</p> <p style="text-align: center;"><img alt="" src="../../em-induction/laminate.svg" style="width: 350px; height: 167px;"></p> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> </div> </div> </div> <div class="panel panel-has-colored-body panel-green"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test Yourself</p> </div> </div> <div class="panel-body"> <div> <p><em>Use quizzes to practise application of theory.</em></p> <br><a class="btn btn-primary btn-block text-center" data-toggle="modal" href="#c8b93d9f"><i class="fa fa-play"></i> START QUIZ!</a><div class="modal fade modal-slide-quiz" id="c8b93d9f"> <div class="modal-dialog" style="width: 95vw; max-width: 960px"> <div class="modal-content"> <div class="modal-header slide-quiz-title"> <h4 class="modal-title" style="width: 100%;"> Electromagnetic induction <strong class="q-number pull-right"> <span class="counter">1</span>/<span class="total">1</span> </strong> </h4> </div> <div class="modal-body p-xs-3"> <div class="slide-quiz" data-stats="6-350-1142" style="opacity: 0"> <div class="exercise shadow-bottom"><div class="q-question"><p>A conducting ring passes through the magnetic field as shown.</p><p style="text-align: center;"><img alt="" height="254" class="gifffer" data-gifffer="/media/physics/electricity/emi1.gif" width="302"></p><p>The induced current will flow:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>clockwise</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>anti-clockwise then clockwise</span></label> </p><p><label class="radio"> <input type="radio"> <span>clockwise the anti-clokwise</span></label> </p><p><label class="radio"> <input type="radio"> <span>anti-clockwise</span></label> </p></div><div class="q-explanation"><p>We can use Fleming's right hand rule to determine the directions of the current.</p><p>As the bottom of the ring enters the field it is moving downwards and the field is into the page. Conventional current is to the right and so flows anti-clockwise.</p><p style="text-align: center;"><img alt="" height="222" src="../../electricity/screenshot-2019-09-21-at-10.02.34.png" width="212"></p><p>As the top of the ring leaves the field it is moving downwards and the field is into the page. Conventional current is to the right and so flows clockwise.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A conducting ring passes through a magnetic field as shown.</p><p style="text-align: center;"><img alt="" class="gifffer" data-gifffer="/media/physics/electricity/emi2.gif" style="width: 250px; height: 159px;"></p><p>The induced current will flow:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>clockwise</span></label> </p><p><label class="radio"> <input type="radio"> <span>clockwise then anti-clockwise</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>anti-clockwise then clockwise</span></label> </p><p><label class="radio"> <input type="radio"> <span>anti-clockwise</span></label> </p></div><div class="q-explanation"><p>We can use Fleming's right hand rule to determine the directions of the current.</p><p>As the right of the ring enters the field it is moving to the right and the field is into the page. Conventional current is upward and so flows anti-clockwise.</p><p style="text-align: center;"><img alt="" height="180" src="../../electricity/screenshot-2019-09-21-at-10.14.50.png" width="185"></p><p>As the left of the ring leaves the field it is moving to the right and the field is into the page. Conventional current is upward and so flows clockwise.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A conducting ring passes through a uniform magnetic field at constant speed as shown.</p><p style="text-align: center;"><img alt="" height="215" class="gifffer" data-gifffer="/media/physics/electricity/emi3.gif" width="208"></p><p>What can be said about the induced current?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Induced current flows clockwise</span></label> </p><p><label class="radio"> <input type="radio"> <span>Induced current flows anti-clockwise</span></label> </p><p><label class="radio"> <input type="radio"> <span>Induced current flows anti-clockwise then clockwise</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>No current will flow</span></label> </p></div><div class="q-explanation"><p>There is no change in flux as the magnetic field and the velocity of the ring are constant. Therefore there is no induced EMF (depends on rate of change of flux) and so no induced current.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A metal bar passes through a magnetic field as shown.</p><p style="text-align: center;"><img alt="" height="225" class="gifffer" data-gifffer="/media/physics/electricity/emi4.gif" width="188"></p><p>Which statement is true about the charge at the ends of the bar?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Both ends are neutral</span></label> </p><p><label class="radio"> <input type="radio"> <span>Both ends are +ve</span></label> </p><p><label class="radio"> <input type="radio"> <span>Left is +ve and right is -ve</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Left is -ve and right is +ve</span></label> </p></div><div class="q-explanation"><p>In this case we use Fleming's left hand rule to determine the direction of force on each free electron. The bar containing the electrons moves downwards, so conventional current is upward. The field is into the page. The force on an electron is to the left and so the left side of the bar will become negative (leaving a net positive charge on the right)</p><p style="text-align: center;"><img alt="" height="213" src="../../electricity/screenshot-2019-09-21-at-10.49.22.png" width="170"></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A conductor of length <span class="math-tex">\(l\)</span> travels with velocity <span class="math-tex">\(v\)</span> through a magnetic field of flux density <span class="math-tex">\(B\)</span>.</p><p style="text-align: center;"><img alt="" height="188" class="gifffer" data-gifffer="/media/physics/electricity/emi4.gif" width="157"></p><p>What is the magnitude of the induced EMF?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(Bv\over l\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(BL\over v\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(0\)</span></span></label></p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(Blv\)</span></span></label></p></div><div class="q-explanation"><p>The EMF is equal to the flux cut per unit time: <span class="math-tex">\(Blvt\over t\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Two magnets are dropped through two coils as shown.</p><p style="text-align: center;"><img alt="" height="280" src="../../electricity/screenshot-2019-09-21-at-14.05.04.png" width="309"></p><p>As the magnets enter the coils, the EMF induced reaches a maximum. If the maximum value across the left hand coil is <span class="math-tex">\(\varepsilon\)</span>, what is the EMF across the right hand coil?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(\varepsilon\)</span></span></label></p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(2\varepsilon\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(4\varepsilon\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(8\varepsilon\)</span></span></label> </p></div><div class="q-explanation"><p>All factors including field strength and speed are the same except for the number of turns on the coil:</p><p style="text-align: center;"><span class="math-tex">\(\varepsilon\propto N\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Two magnets are dropped through two short circuited coils as shown.</p><p style="text-align: center;"><img alt="" height="280" src="../../electricity/screenshot-2019-09-21-at-14.05.04.png" width="309"></p><p>As the magnets enter the coils, the current induced reaches a maximum. If the maximum current in the left hand coil is <span class="math-tex">\(I\)</span>, what is the current in the right hand coil?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(2I\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(8I\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(I\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(4I\)</span></span></label> </p></div><div class="q-explanation"><p>The size of the induced EMF is double for the right hand coil as the number of turns has doubled. However, the right hand coil is twice the length and so will have double the resistance.</p><p>Since <span class="math-tex">\(I={V\over R}\)</span>, the current is the same.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Two magnets are dropped into two identical coils. The magnet on the left is twice as high above the coil as the one on the right.</p><p style="text-align: center;"><img alt="" height="216" src="../../electricity/screenshot-2019-09-21-at-14.05.041.png" width="240"></p><p>As the magnets enter the coils, the EMF induced reaches a maximum. If the maximum value in the left hand coil is <span class="math-tex">\(E\)</span>, what is the maximum EMF in the right hand coil?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(E\over2\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(E\over 8\)</span></span></label></p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(E\over √2\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(E\over4\)</span></span></label></p></div><div class="q-explanation"><p><span class="math-tex">\(E\)</span> is proportional to the magnet's velocity when it enters the coil.</p><p>Using equations of uniformly accelerated motion or conservation of mechanical energy, <span class="math-tex">\(v^2 = 2as\)</span>:</p><p style="text-align: center;"><span class="math-tex">\(v\propto \sqrt s \)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A square coil with <span class="math-tex">\(N\)</span> turns and area <span class="math-tex">\(A\)</span> is positioned in a field of flux density <span class="math-tex">\(B\)</span> as shown:</p><p style="text-align: center;"><img alt="" height="193" src="../../electricity/screenshot-2019-09-21-at-14.53.15.png" width="172"></p><p>The flux enclosed is:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(BAN\tan a\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(BAN\sin a\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(BAN\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(BAN\cos a\)</span></span></label> </p></div><div class="q-explanation"><p>The component of the flux denity perpendicular to the plane of the coil is <span class="math-tex">\(B\sin a\)</span>.</p><p>It is always worth scrutinising angles that you are given in diagrams rather than blindly using <span class="math-tex">\(\sin\)</span> or <span class="math-tex">\(\cos\)</span> as suggested in the Data Booklet.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following is not a unit of magnetic flux density?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>NC<sup>-1</sup>m<sup>-1</sup>s</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>NCms<sup>-1</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>T</span></label> </p><p><label class="radio"> <input type="radio"> <span>NA<sup>-1</sup>m<sup>-1</sup></span></label> </p></div><div class="q-explanation"><p>T is the SI derived unit of magnetic flux density.</p><p>NA<sup>-1</sup>m<sup>-1</sup> is from the equation <span class="math-tex">\(F=BIl\)</span> for the magnetic force exerted on a current-carrying conductor.</p><p>NC<sup>-1</sup>m<sup>-1</sup>s is from the equation <span class="math-tex">\(F=Bqv\)</span> for the magnetic force exerted on a charge. Therefore, NCms<sup>-1</sup> cannot also be accurate.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The graph represents the changing flux density in a coil.</p><p style="text-align: center;"><img alt="" height="173" src="../../electricity/screenshot-2019-09-22-at-07.55.56.png" width="238"></p><p>If the EMF induced in the coil is 2V when the flux changes at the rate of line A, what will the EMF be if the flux changes at the rate of line B?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(V\over2\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(V\over 3\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(V\over \sqrt3\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(V\over 6\)</span></span></label> </p></div><div class="q-explanation"><p>EMF is proportional to the rate of change of flux. The gradient of B is <span class="math-tex">\(1\over 6\)</span> of the gradient of A, so EMF will be <span class="math-tex">\({1\over 6} \times 2\)</span> .</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Two magnets are dropped into two coils. The magnet on the left is twice as high above the coil as the one on the right.</p><p style="text-align: center;"><img alt="" height="241" src="../../electricity/screenshot-2019-09-21-at-14.05.042.png" width="265"></p><p>As the left magnet passes into its coil the current induced reaches a maximum value <span class="math-tex">\(I\)</span>. What is the maximum current in the right hand coil?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(I\over \sqrt2\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(0\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(I\over 2\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(I\over 4\)</span></span></label> </p></div><div class="q-explanation"><p>A close inspection of the diagram reveals that the right hand coil is open - no current can flow!</p><p>Had the circuit been shorted like the first, the size of the maximum current would have been <span class="math-tex">\(I\over \sqrt2\)</span>.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> 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