Question

The resistance of a circular coil of $50$ turns and $10\, cm$ 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, $n=50,\, r=\frac{10}{2}=5 cm =5 \times 10^{-2}\, m$
$R=5 \Omega$ and $H=0.314$ gauss
For circular coil, $H=\frac{\mu_{0}}{4 \pi} \frac{2 \pi n i}{r}$
or $i=\frac{4 \pi}{\mu_{0}} \frac{ HH }{2 \pi n}$
$=\frac{5 \times 10^{-2} \times 0.314 \times 10^{-4}}{10^{-6} \times 2 \times 3.14 \times 50}$
$=5 \times 10^{-2} A$
$V = iR$
$=5 \times 10^{-2} \times 5=025\, V$
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.