Question

A square loop of wire of each side 50 cm is
kept so that its plane makes angle o
s with a uniform mag. field of induction
0.1 tesla. The magnetic field is withdrawn
in 0.1 sec. It is found that the induced
emf across the loop is 125 mV. the angle O is
A) 30° B) 45° C) 60° D) 90°​

Answer 1

A square loop of wire of each side 50 cm is kept so that its plane makes angle θ with a uniform magnetic field of induction 0.1 T .The magnetic field is withdrawn in 0.1 sec. it is found that the induced emf across the loop is 125 mV.

We have to find the angle θ.

area of square loop , A = (edge length of square loop)²

= (50 cm)²

= 2500 cm²

= 2500 × 10¯⁴ m²

= 0.25 m²

strength of magnetic field, B = 0.1 T

time, t = 0.1 sec

induced emf = 125 mV = 0.125 V

we know, “ the induced emf is the negative of the rate of change of magnetic flux through the surface. ”

i.e., emf = -Φ/t = -BAcos(90 +θ)/t

[ see figure, a/c to question θ is the angle between magnetic field and plane so the (90 + θ) will be the angle between magnetic field and area vector ]

⇒0.125 = -1 × 0.25 × cos(90 + θ)/0.1

⇒0.125/0.25 = 1/2 = -cos(90 + θ) = sinθ

⇒θ = 30°

Therefore the angle plane makes an angle 30° with the magnetic field.