Question

A circular coil of radius $8.8$ cm has $10$ turns, a current of $1.0$ A is passing through it. Find the resultant field at its centre when the plane of the coil is vertical and perpendicular to the magnetic meridian. The horizontal component of the earth’s magnetic induction $=B_H=0.4\times {10}^{-4}T,B_V=0$

• A. $0.31\times {10}^{-4}T$ towards south
• B. $1.31\times {10}^{-4}T$ towards south
• C. $0.31\times {10}^{-4}T$ towards north
• D. $1.31\times {10}^{-4}T$ towards north

Answer 1

The correct option is A $0.31\times {10}^{-4}T$ towards south

Given, radius of coil $=8.8cm$.
$N=10$ turns, $B_H=0.4\times {10}^{-4}T$,$i=1A$.
$B_{centre}=\frac{{\mu }_0}{2}\frac{NI}{r}=\frac{4\pi \times {10}^{-7}\times 10\times 1.0}{2\times 8.8\times {10}^{-2}}=0.71\times {10}^{-4}T$

$B_{net}=B_{centre}\pm B_H=0.71\times {10}^{-4}\pm 0.4\times {10}^{-4}=1.11\times {10}^{-4}T$ (Towards North)
$B=0.31\times {10}^{-4}T$ towards south.