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Gauss's Law for Magnetism

Q: Use Lenz's law to determine the direction of induced current in the situations described by Fig.(a) A wire of irregular shape turning into a circular shape;(b) A circular loop being deformed into a narrow straight wire.

According to the Lenz's law, the current induced will oppose the change in flux through the loop.(a) Since the area of loop is increasing with time, the flux through it will increase and hence the current induced will be such that its magnetic field produced is in the opposite direction.Hence according to right hand rule, the current will be in clockwise direction(b) Since the area of loop is decreasing with time, the flux through it will decrease and hence the current induced will be such that its magnetic field produced is in the same direction to maintain the flux right hand rule, Direction of current, - adcb

Q: Two identical coaxial circular loops carry a current i each circulating in the same direction. If the loops approach each other, you will observe that the current in

As the two identical coaxial circular loops carry a current i each circulating in the same direction, their flux add each other. When the two are brought closer, the total fluc linkage increase. By Lenz's a law a current will be induced in both the loops such that it opposes this change in flux. As the flux increases, the current in each loop will decreases so as to decrease the increasing flux.

Q: A conducting wire frame is placed in a magnetic field which is directed into the plane of the paper (figure). The magnetic field is increasing at constant rate. The directions of induced currents in wires \$AB\$ and \$CD\$ are

Consider the surface facing upwards having points A, B, C and D in anticlockwise direction when viewing towards the surface from above. Let this be \$ \vec S \$flux = \$ \vec S \times \vec B \$As the flux increases, by Lenz's Law the induced current will be in the direction C B A D C, so as to oppose the increase in flux.

Q: A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be:

\$P=\frac{\varepsilon ^2}{R}\$where,\$\varepsilon =induced emf=-\frac{d\phi}{dt}\$\$\phi=NBA\$\$\varepsilon =NA\frac{dB}{dt}\$also, \$R\propto \frac{l}{r^2}\$where ,R=resistance and r=radius of wire\$P\propto \frac{N^2r^2}{l^2}\$\$\frac{P_1}{P_2}=1\$\$P_1=P_2\$This means if the number of turns were quadrupled and the wire radius halved, the electrical power dissipated would be remains same.

Q: A square loop of wire, side length 10 cm is placed at angle of \$45^{o}\$ with a magnetic field that changes uniformly from 0.1 T to zero in 0.7 seconds. The induced current in the loop ( its resistance is \$1\Omega\$) is

\$L=10cm=10^{-2}m\$ ,\$\theta=45\$Initial magnetic field \$B_i=0.1T\$Final magnetic field \$B_f=0T\$\$\Delta t=0.7s\$ ,\$R=1 \Omega\$Now \$\phi_B=BA \cos\theta=\dfrac{ { 10 }^{ -3 } }{ \sqrt { 2 } } \$wB\${ \phi }_{ { B }_{ f } }=0\quad wb\$\$\varepsilon =\dfrac{\left| \Delta { \phi }_{ B } \right| }{ \Delta t } =\dfrac{ { 10 }^{ -3 } }{ \sqrt { 2 } } \times \dfrac{ 1 }{ 0.7 } =1mV\$\$I=\dfrac{ \varepsilon }{ R } =\dfrac{ 1mV }{ 1\Omega } =1mA\$

Q: A metallic rod of mass per unit length \$0.5\$ kg \$m^{-1}\$ is lying horizontally on a smooth inclined plane which makes an angle of \$30^o\$ with the horizontal. The rod is not allowed to slide down by a flowing a current through it when a magnetic field of induction \$0.25\$T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is?

\$B=0.25 T\$ \$\displaystyle\frac{m}{l}=0.5 \dfrac{kg}{m}\$ \$\theta =30^o\$ \$F=Bil\$ \$F\cos 30\$ balances \$mg \sin 30\$ \$\therefore (Bil) cos30^{o} =mg\sin 30\$ \$\Rightarrow i=\displaystyle\frac{m}{l}\frac{g}{B}\displaystyle\frac{\sin 30}{\cos 30}=\frac{0.5\times 9.8}{0.25\times 866}\times \frac{1}{2}=11.32A\$

Q: Lenz's law is in accordance with the law of

Lenz's law states - The direction of current induced in a conductor by a changing magnetic field due to Faraday's law of induction will be such that it will create a field that opposes the change that produced it.The induced EMF produces a current that opposes the change in flux and hence energy. As the change starts, induction opposes and slows the change. If the induced EMF were in the same direction as the change in flux, then that would give us free energy from no apparent source violating conservation of energy. Thus Lenz's law is in accordance with the law of conservation of energy.

Q: A copper ring is held horizontally and a bar magnet is dropped through the ring with its length along the axis of the ring. Then, the acceleration of the falling magnet (neglect air resistance) will be

As the magnet falls the flux through the ring changes. Regardless whether the flux increases or decreases, by Lenz's law a current is induced in the ring that opposes the change in flux. This induced current induced a magnetic field that interacts with the falling magnet and decreases the acceleration of fall below \$g\$.

Q: In Figure , a metal rod is forced to move with constant velocity \$\overrightarrow {v}\$ along two parallel metal rails, connected with a strip of metal at one end. A magnetic field of magnitude B = 0.350 T points out of the page. (a) If the rails are separated by L = 25.0 cm and the speed of the rod is 55.0 cm/s, what emf is generated? (b) If the rod has a resistance of \$18.0\Omega \$ and the rails and connector have negligible resistance, what is the current in the rod?(c) At what rate is energy being transferred to thermal energy?

(a) Equation \$\mathscr{E}=\dfrac{d\Phi_B}{dt}=\dfrac{d}{dt}BLx=BL\dfrac{dx}{dt}=BLv\$ leads to\$\varepsilon=B L v=(0.350\mathrm{T})(0.250 \mathrm{m})(0.55 \mathrm{m} / \mathrm{s})=0.0481 \mathrm{V}\$(b) By Ohm's law, the induced current is\$i=0.0481 \mathrm{V} / 18.0\Omega=0.00267 \mathrm{A}\$By Lenz's law, the current is clockwise in Figure.(c) Equation \$P=i^2R\text{ (resistive dissipation) }\$ leads to \$P=i^{2}R=0.000129 \mathrm{W}\$

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