Question

In given figure, a wire loop has been bent so that it has three segments AB (a quarter circle), BC (a square corner) & CA (straight). Here are three choices for a magnetic field through the loop-

(1) $\overset{\rightarrow}{B_{1}} = 3\widehat{i} + 7\widehat{j}–5t\widehat{k}$ (2) $\overset{\rightarrow}{B_{2}} = 5t\widehat{i}–4\widehat{j}–15\widehat{k}$

(3) $\overset{\rightarrow}{B_{3}} = 2\widehat{i}–5t\widehat{j}–12\widehat{k}$

where B is in milli tesla and t is in second if the induced current in the loop due to $\overset{\rightarrow}{B_{1}},\overset{\rightarrow}{B_{2}},\overset{\rightarrow}{B_{3}}$ are i₁, i₂, i₃ respectively, then -

• A. i₁ > i₂ > i₃
• B. i₂ > i₁ > i₃
• C. i₃ > i₂ > i₁
• D. i₁ = i₂ = i₃

Answer 1

Answer: (B)

i₂ > i₁ > i₃

Answer 2

i₁ = $\frac{d\varphi_{1}}{dt}\frac{1}{R} = \frac{1}{R} \times \frac{d}{dt}(B_{1} \times Area) = \frac{d}{dt}\left\lbrack 5t \times \frac{\pi a^{2}}{4} \right\rbrack \times \frac{1}{R} = \frac{5\pi a^{2}}{4R}$

i₂ = $\frac{d\varphi_{2}}{dt} \times \frac{1}{R} = \frac{d\lbrack 5a^{2}t\rbrack}{dt} \times \frac{1}{R} = \frac{5a^{2}}{R}$

i₃ = $\frac{d\varphi_{3}}{dt} \times \frac{1}{R} = \frac{d\lbrack 0.5a^{2}\rbrack}{dt} \times \frac{1}{R} = \frac{a^{2}}{2R}$

\ i₂ > i₁ > i₃