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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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A circular coil of $N=5$ turns with radius $r=1.73 m$ is subjected to a varying magnetic field that changes uniformly from $2.18 T$ to $12.7 T$ in an interval of $180 \text{seconds}$. The axis of the solenoid makes an angle of $27 ^{\circ}$ to the magnetic field. Find the induced emf.

(A) 1.22 V

(B) 3.67 V

(C) 4.9 V

(D) 2.45 V

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Given $N = 5, r = 1.73m, B_1 = 2.18T, B_2 = 12.7 T, \theta = 27$
Induced emf $\varepsilon = \large\frac{d\Phi}{dt} = \large\frac{d NBA \cos \theta }{dt} =$$ NA\cos \theta $$\large\frac{B_2-B_1}{dt}$
$\varepsilon = 5 \times \pi \times 1.73^2 \times \cos 27 \times \large\frac{12.7-2.18}{180}$$ = 2.45V$
answered Mar 12, 2014 by balaji.thirumalai

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