Question

A plane loop is shaped in the form as shown in figure with radii $a=20\text{cm}$ and $b=10\text{cm}$ and is placed in a uniform time varying magnetic field $B=B_0\sin \left(\omega t\right)$ where $B_0=10\text{m T}$ and $\omega =100{\text{rad s}}^{-1}$. The amplitude of the induced current in the loop is $n\text{A}.$ The value of $n$ is (integer only)


[Resistance per unit length $\lambda =50\times {10}^{-3}\Omega {\text{m}}^{-1}$ and inductance of the loop is negligible.]

• A. 1
• B. 1.00
• C. 1.0

Answer 1

Answer: (A), (B), (C)

Given, $a=20\text{cm};b=10\text{cm}B=B_0\sin \left(\omega t\right);B_0=10\text{mT};\omega =100{\text{rad s}}^{-1}\lambda =50\times {10}^{-3}\Omega {\text{m}}^{-1}$

Instantaneous flux through loop is,

$\varphi =\pi a^{2}B\cos 0^{\circ }+\pi b^{2}B\cos {180}^{\circ }$

$\Rightarrow \varphi =\pi (a^2-b^2)B$

$\Rightarrow \varphi =\pi (a^2-b^2)B_0\sin \left(\omega t\right)$

Differentiating w.r.t. time, we get

$e=\left|\frac{d\varphi }{dt}\right|=\pi (a^2-b^2)B_0\omega \cos \left(\omega t\right)$

$\Rightarrow I=\frac{\pi (a^2-b^2)B_0\omega \cos \left(\omega t\right)}{R}$

$I_{max}=I_0=\frac{\pi (a^2-b^2)B_0\omega }{2\pi (a+b)\lambda }[\because \lambda =\text{resistance per unit length}]$

$I_{max}=I_0=\frac{1}{2\lambda }(a-b)B_0\omega$

$I_{max}=I_0=\frac{1}{2}(20-10)\times {10}^{-2}\times 10\times {10}^{-3}\times \frac{100}{50\times {10}^{-3}}$

$I_{max}=I_0=1\text{A}$