Question

A square loop of side ‘a’ and resistance R is placed in a transverse uniform magnetic field B. If it suddenly changes into circular form in time t then magnitude of induced charge will be

• A. $\frac { B a ^ { 2 } } { R } ( 4 \pi - 1 )$
• B. $\frac { B a ^ { 2 } } { R } \left( 1 - \frac { 1 } { 4 \pi } \right)$
• C. $\frac { B a ^ { 2 } } { R } \left( \frac { 1 } { 4 \pi } - 1 \right)$
• D. $\frac { B a ^ { 2 } } { R } \left( \frac { 4 } { \pi } - 1 \right)$

Answer 1

Answer: (D)

$\frac { B a ^ { 2 } } { R } \left( \frac { 4 } { \pi } - 1 \right)$

Answer 2

Initially It’s area A₁ = a² ; and flux linked φ₁ = BA₁

Finally It’s area A₂ = πr² $= \pi \left( \frac { 2 a } { \pi } \right) ^ { 2 } = \frac { 4 a ^ { 2 } } { \pi }$ and flux

linked φ₂ = BA­₂­

Induced emf $| e | = \frac { \Delta \phi } { \Delta t } = \frac { \phi _ { 2 } - \phi _ { 1 } } { \Delta t } = \frac { B \left( A _ { 2 } - A _ { 1 } \right) } { \Delta t }$

$= \frac { B a ^ { 2 } } { t } \left( \frac { 4 } { \pi } - 1 \right)$ so induced charged

$= \frac { B a ^ { 2 } } { R } \left( \frac { 4 } { \pi } - 1 \right)$