Question

A uniform magnetic field, fills a cylindrical volume of radius 10 cm as shown is figure. If the magnitude of magnetic field increases in the direction of field at a constant rate of 10T/s then the emf induced in the rod AB is:

• A. $24mV$
• B. $96mV$
• C. $48mV$
• D. $zero$

Answer 1

Answer: (C)

$48mV$

The electric field at a distance r from the center is given by
$E=\frac{\pi r^{2}dB/dt}{2\pi r}=\frac{r}{2}\frac{dB}{dt}$
The field is directed perpendicular to the radial lines
If h is the distance from the center of the cylinder to the center of the rod we evaluate
$ϵ=\frac{dB}{dt}\int \frac{r}{2}\frac{h}{r}dx$
$=\frac{dB}{dt}\frac{L}{2}h$
But we have
$h^2=R^2-(\frac{L}{2}{)}^2$
Thus
$ϵ=\frac{dB}{dt}\frac{L}{2}\sqrt{R^2-(\frac{L}{2}{)}^2}$
Substituting the values we get
$ϵ=10\times \frac{0.12}{2}\times \sqrt{{0.1}^2-(\frac{0.12}{2}{)}^2}$
Thus we get
$ϵ=48\times {10}^{-3}V=48mV$