Question

A magnetic field induction is changing in magnitude in a region at a constant rate $\frac{dB}{dt}$. A given mass $m$ of copper drawn into a wire and formed into a loop is placed perpendicular to the field. If the values of specific resistance and density of copper are $\rho$ and $\sigma$ respectively, then the current in the loop is given by :

• A. $\frac{4\pi m}{\rho \sigma }\frac{dB}{dt}$
• B. $\frac{m}{4\pi \rho \sigma }\frac{dB}{dt}$
• C. $\frac{m}{\rho \sigma }\frac{dB}{dt}$
• D. $\frac{2\pi m}{\rho \sigma }\frac{dB}{dt}$

Answer 1

The correct option is B $\frac{m}{4\pi \rho \sigma }\frac{dB}{dt}$

Radius of wire $=r$ Radius of loop$=R_1$
Cross-sectional area of wire $=\pi r^2$ Length of wire $=2\pi R_1$
Total resistance of wire $=R=\rho \times \frac{2\pi R_1}{\pi r^2}$
Mass$=m,density=\sigma$
$\sigma \times \pi r^2\times 2\pi R_1=m$
$\therefore R=\rho \times (2\pi R_1{)}^2\frac{\sigma }{m}$
$emf=\frac{dB}{dt}\times \pi R_1^2$
Current $=\frac{dB}{dt}\frac{\pi R_1^2}{R}$
$=\frac{m}{\sigma \rho }\frac{\pi R_1^2}{4{\pi }^2{R}_1^2}\frac{dB}{dt}$
$=\frac{m}{4\pi \sigma \rho }\frac{dB}{dt}$