A flat circular coil of n turns, area A and resistance R is placed in a uniform magnetic field B. The plane of coil is initially perpendicular to B. When the coil is rotated through an angle of 180 o about one of its diameter, a charge Q 1 flows through the coil. When the same coil after being brought to its initial position, is rotated through an angle of 360 o about the same axis a charge Q 2 flows through it. Then Q 2 /Q 1 A 1 B 2 C 1/2 D 0

A flat circular coil of $n$ turns, area $A$ and resistance $R$ is placed in a uniform magnetic field $B$ . The plane of coil is initially perpendicular to $B$ . When the coil is rotated through an angle of $180^{o}$ about one of its diameter, a charge $Q_{1}$ flows through the coil. When the same coil after being brought to its initial position, is rotated through an angle of $360^{o}$ about the same axis a charge $Q_{2}$ flows through it. Then $Q_{2} / Q_{1}$

A

$1$

B

$2$

C

$1 / 2$

D

$0$

Hard

Solution

Verified by GMS

Correct option is D)

Net charge flowing through the coil is given by $Q = \frac{Δ ϕ}{R}$    where $R$ is the resistance

Initially the plane of ring is perpendicular to $B$ , i.e Area vector $A$ is parallel to $B$
where $A$ is the area of the coil.

$∴$   initial flux    $ϕ_{i} = A . B = A B$

Case 1) : Coil is rotated by $18 0^{o}$    i.e    $A$ $∥$     $- B$

$∴$ Final flux    $ϕ_{f} = - A B$

$⟹ ∣ Δ ϕ ∣ = ∣ ϕ_{f} - ϕ_{i} ∣ = 2 A B$

Thus $Q_{1} = \frac{2 A B}{R}$    .............(1)

Case 2): Coil is rotated by $36 0^{o}$ , i.e $A$    $∥$    $B$

Thus final flux    $ϕ_{f} = A B$

$∴ Δ ϕ = ϕ_{f} - ϕ_{i} = 0$

Hence    $Q_{2} = \frac{Δ ϕ}{R} = 0$

$⟹ \frac{Q _{2}}{Q _{1}} = 0$

Solve any question of Electromagnetic Induction with:-

Patterns of problems

Getting Info...

Copy