Question

The resistance of a circular coil of $50$ turns and $10cm$ diameter is $5\Omega$. What must be the potential difference across the ends of the coil so as to nullify the earth's magnetic field ($H=0.314$ gauss ) at the centre of the coil? How should the coil be placed to achieve this result?

• A. 0.5 V with plane of coil normal to die magnetic meridian
• B. 0.5 V with plane of coil in the magnetic meridian
• C. 0.25 V with plane of coil normal to the magnetic meridian
• D. 0.25 V with plane of coil in the magnetic meridian

Answer 1

Answer: (C)

Given, $n=50,r=\frac{10}{2}=5cm=5\times 10^{-2}m$
$R=5\Omega$ and $H=0.314$ gauss
For circular coil, $H=\frac{{\mu }_0}{4\pi }\frac{2\pi ni}{r}$
or $i=\frac{4\pi }{{\mu }_0}\frac{HH}{2\pi n}$
$=\frac{5\times 10^{-2}\times 0.314\times 10^{-4}}{10^{-6}\times 2\times 3.14\times 50}$
$=5\times 10^{-2}A$
$V=iR$
$=5\times 10^{-2}\times 5=025V$
Magnetic field produced due to the coil is normal to the plane of coil so the plane of the coil should be normal to the magnetic meridian.