Question

A vertical circular coil of radius $0.1m$ and having $10$ turns carries a steady current. When the plane of the coil is normal to the magnetic meridian, a neutral point is observed at the centre of the coil. If $B_H=0.314\times {10}^{-4}T$, the current in the coil is :

• A. $0.5A$
• B. $0.25A$
• C. $2A$
• D. $1A$

Answer 1

The correct option is A $0.5A$

The magnetic field at the center of a current carrying coil is given by the formula,

$B=\frac{{\mu }_{0}nI}{2R}$

Hence, for the net magnetic field to be zero, this field must cancel the horizontal component of earth.

Hence,

$\frac{{\mu }_{0}nI}{2R}=B_H$

$I=\frac{2R{B}_H}{n{\mu }_0}=\frac{2\left(0.1\right)(0.314\times {10}^{-4})}{\left(10\right)4\left(3.142\right)\times {10}^{-7}}=0.5A$