Electromagnetic Induction

Q: A 800 turn coil of effective area 0.05 m^2 is kept perpendicular to a magnetic field 5 \times 10^{-5} T . When the plane of the coil is rotated by 90^o around any of its coplanar axis in 0.1 s, the emf induced in the coil will be :

Magnetic field B = 5 \times 10^{-5} T Number of turns in coil N = 800 Area of coil A = 0.05 m^2 Time taken to rotate \Delta t = 0.1 s Initial angle \theta_1 = 0^o Final angle \theta_2 = 90^o change in magnetic flux \Delta \phi = NBA cos 90^o - BAcos 0^o = - NBA =-800 \times 5 \times 10^{-5} \times 0.05 = -2 \times 10^{-3} weber e = - \dfrac{\Delta \phi}{\Delta t} = \dfrac{-(-)2 \times 10^{-3} Wb}{0.1 s} = 0.02 V

Q: In which of the following devices, the eddy current effect is not used ?

Electric heater does not involve Eddy currents. It uses Joule's heating effect and hence will be the correct answer

Q: The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance.

Energy stored in a Inductor is given as U = \dfrac{1}{2} LI^2 L = \dfrac{2U}{I^2} L = \dfrac{2 \times 25 \times 10^{-3}}{(60 \times 10^{-3})^2} L = 13.89 H

Q: A long solenoid of diameter 0.1\ m has 2 \times {10^4} turns per metre.At the centre of the solenoid, a coil of 100 turns and radius 0.01\ m is placed with its axis coinciding with the solenoid axis.The current in the solenoid reduces at a constant rate to 0\ A from 4\ A in 0.05\ s . If the resistance of the coil is 10 \ {\pi ^2}\Omega , the total charge flowing through the coil during this time is.

Given,Number of turns, n=100 Radius, r=0.01\,m Resistance, R=10\pi^2 \Omega As we know, \epsilon=-N\dfrac{d\phi}{dt} =\dfrac{\epsilon}{R}=-\dfrac NR\dfrac{d\phi}{dt} , \Delta I=-\dfrac NR\dfrac{d\phi}{dt} \dfrac{\Delta}{\Delta t}=-\dfrac NR\dfrac{\Delta\phi}{\Delta t}\implies \Delta q=-[\dfrac NR(\dfrac{\Delta \phi}{\Delta t})]\Delta t - ve sign shoes that induced emf opposes the change in flux. \Delta q=\dfrac{\mu_0 ni\pi r^2}{R} \Delta q=\dfrac{4\pi\times 10^{-7}\times 100\times 4\times \pi\times (0.01)^2}{10\pi^2}=32\mu C

Q: A long solenoid has 1000\ turns. When a current of 4\ A flows through it, the magnetic flux linked with each turn of the solenoid is 4\times 10^{-3} Wb. The self-inductance of the solenoid is:

Flux linked with each turn =4\times 10^{-3}\ Wb \Rightarrow Total flux linked =1000\times 4\times 10^{-3}\ Wb \Rightarrow \phi_{total}=4 \Rightarrow LI=4 \Rightarrow L=1 \ H