A square loop of wire of each side 50 cm is kept, so that its plane makes an angle 9 with a uniform magnetic field of induction 1 T. The magnetic field is withdrawn in 0.1 s. It is found that the induced emf across the loop is 125 mV. The angle 0 is

Question

A square loop of wire of each side 50 cm is kept, so that its plane makes an angle 9 with a uniform magnetic field of induction 1 T. The magnetic field is withdrawn in 0.1 s. It is found that the induced emf across the loop is 125 mV. The angle 0 is (A)  90°  (B)  60°  (C)  45°  (D)  30°

Tardigrade Admin · Accepted Answer

e=nAB sin (90o-θ ) =nAB cos θ ∴ 125× 10-3=1× 0.5× 0.5× 1× cos θ cos θ =(1/2) θ =60o

Q. A square loop of wire of each side 50 cm is kept, so that its plane makes an angle 9 with a uniform magnetic field of induction 1 T. The magnetic field is withdrawn in 0.1 s. It is found that the induced emf across the loop is 125 mV. The angle 0 is

$90^{\circ}$

$60^{\circ}$

$45^{\circ}$

$30^{\circ}$

$e = n A B sin (90^{o} - θ)$ $= n A B cos θ$ $∴$ $125 \times 10^{- 3} = 1 \times 0.5 \times 0.5 \times 1 \times cos θ$ $cos θ = \frac{1}{2}$ $θ = 60^{o}$