Q143 / 180 · KCET 2013 · Physics

A rectangular coil of 100 turns and size $0.1 \mathrm{~m} \mathrm{} \times 0.05 \mathrm{~m}$ is placed perpendicular to a magnetic field of $0.1 \mathrm{~T}$. If the field drops to $0.05 \mathrm{~T}$ in $0.05 \mathrm{~s}$, the magnitude of the e.m.f. iniduced in the coil is

  • A
    $\sqrt{2}$
  • B
    $\sqrt{3}$
  • C
    $\sqrt{0.6}$
  • Correct answer
    D
    $\sqrt{6}$

Solution

Given, $n=100$ turns, $A=0.1 \times 0.05 \mathrm{~m}^{2}$, $B_{1}=0.1 \mathrm{~T}, \quad B_{2}=0.05 \mathrm{~T}, d t=0.05 \mathrm{~s}$ We know that, $e=\left|\frac{-d \phi}{d t}\right|=\frac{d}{d t}(n B A \cos \theta)=\frac{n A d B \cos \theta}{d t}$ Here, $\quad \theta=0^{\circ}$ $\begin{aligned} &\therefore \quad e=\frac{n A d B}{d t}=\frac{100 \times 0.1 \times 0.05 \times(0.1-0.05)}{0.05} \\ &\Rightarrow e=0.5 \mathrm{~V} \end{aligned}$

Hence, the correct answer is option: 'D'.

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