 # A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of a metal wire. Question:

A uniform magnetic field $B$ exists in a direction perpendicular to the plane of a square loop made of a metal wire. The wire has a diameter of $4 \mathrm{~mm}$ and a total length of $30 \mathrm{~cm}$. The magnetic field changes with time at a steady rate $d B / d t=$ $0.032 \mathrm{Ts}^{-1}$. The induced current in the loop is close to (Resistivity of the metal wire is $1.23 \times 10^{-8} \Omega \mathrm{m}$ )

1. (1) $0.43 \mathrm{~A}$

2. (2) $0.61 \mathrm{~A}$

3. (3) $0.34 \mathrm{~A}$

4. (4) $0.53 \mathrm{~A}$

Correct Option: , 2

Solution:

(2) Given,

Length of wire, $l=30 \mathrm{~cm}$

Radius of wire, $r=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$

Resistivity of metal wire, $\rho=1.23 \times 10^{-8} \Omega \mathrm{m}$

Emf generated, $|e|=\frac{d \phi}{d t}=\frac{d B}{d t}(A) \quad(\because \phi=$ B.A. $)$

Current, $i=\frac{e}{R}$

But, resistance of wire, $R=\rho \frac{l}{A}$

$\therefore i=\left|\frac{d B}{d t}\right| \frac{(A)^{2}}{\rho l}=\frac{0.032 \times\left\{\pi \times 2 \times 10^{-3}\right\}^{2}}{1.23 \times 10^{-8} \times 0.3}=0.61 \mathrm{~A}$  