Question

A uniform field of induction $B$ is changing in magnitude at a constant rate $dB/dt$. You are given a mass $m$ of copper which is to be drawn into a wire of radius $r$ and formed into a circular loop of radius $R$. Show that the induced current in the loop does not depend on the size of the wire of the loop. Assuming $B$ perpendicular the loop prove that the induced current $i=\frac{m}{4\pi \rho \delta }\frac{dB}{dt}$ where $\rho$, is the resistivity and $d$ is the density of copper.

Answer 1

Given,

Mass of wire $=m$

Density of wire $=d$

Radius of wire $=r$

Radius of wire loop $=p$

Resistivity of wire $=\rho$

Rate change of magnetic field $=\frac{dB}{dt}$

Length of wire $L=2\pi p$

Area of wire $a=\pi r^2$

Area of loop $A=\pi p^2=\frac{4\pi \times \pi p^2}{4\pi }=\frac{2\pi p\times 2\pi p}{4\pi }$

$\Rightarrow A=\frac{L^2}{4\pi }$

Multiply by $\frac{a}{L}$ on both sides,

$\Rightarrow \frac{Aa}{L}=\frac{1}{4\pi }\times aL$ ……. (1)

Volume of wire $V=\frac{m}{d}=aL$ …… (2)

Equating both equation (1) & (2)

$\frac{Aa}{L}=\frac{1}{4\pi }\times \frac{m}{d}$ …… (3)

Resistance of wire is $R$

Emf $E=-\frac{d\varphi }{dt}=-A\frac{dB}{dt}$

$IR=-A\frac{dB}{dt}$

$\Rightarrow I=-\frac{A}{R}\times \frac{dB}{dt}$

$\Rightarrow I=-\frac{A}{\frac{\rho L}{a}}\times \frac{dB}{dt}$

$\Rightarrow I=-\frac{1}{\rho }\times \frac{Aa}{L}\times \frac{dB}{dt}$

From equation (3)

$\Rightarrow I=-\frac{1}{\rho }\times \frac{m}{4\pi d}\times \frac{dB}{dt}$

$\Rightarrow I=-\frac{m}{4\pi \rho d}\times \frac{dB}{dt}$

$m,\rho \&d$ all are constant

Hence proved current is indendent on wire area.