Question

A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle $\theta$ at the centre. The value of the magnetic induction at the centre due to the current in the ring is

• A. proportional to 2 (180${}^o-\theta$)
• B. inversely proportional to r
• C. zero, only if $\theta ={180}^o$
• D. zero for all values of $\theta$

Answer 1

Answer: (D)

zero for all values of $\theta$

Magnetic field at the centre due to current ABC arc is

$B_1=\frac{{\mu }_0}{4\pi }\frac{2\pi I_1}{r}\frac{\theta }{2\pi }$ (Directed upwards)

Magnetic field at the centre due to current in ADC arc is

$B_2=\frac{{\mu }_0}{4\pi }\frac{2\pi I_2}{r}\frac{(2\pi -\theta )}{2\pi }$ (Directed downwards)

Therefore net magnetic field at the centre

$B=\frac{{\mu }_0}{4\pi }\frac{2\pi I_1}{r}\frac{\theta }{2\pi }-\frac{2\pi I_2}{r}\frac{(2\pi -\theta )}{2\pi }$

Also, $I_1=\frac{V}{R_1}=\frac{V}{\rho l_1/A}=\frac{VA}{\rho r\theta }$ and

$I_2=\frac{V}{R_2}=\frac{E}{\rho l_2/A}=\frac{VA}{\rho r(2\pi -\theta )}$

$\therefore B=\frac{{\mu }_0}{4\pi }[\frac{VA}{\rho r\theta }\times \frac{\theta }{r}-\frac{VA}{\rho r(2r-\theta )}\times \frac{(2\pi -\theta )}{r}]=0$