Question

Magnetic field intensity at the centre of coil of $50$ turns, $0.5 \,m$ radius and carrying a current of $2 \,A$, is :

• A. $5 \times 10^{-5} T$
• B. $3 \times 10^{-5} T$
• C. $1.25 \times 10^{-4} T$
• D. $0.5 \times 10^{-5} T$

Answer 1

Answer: (C)

Magnetic field at the centre of a circular coil carrying current $i$ and of radius $a$, having $N$ number of turns is
$\therefore B=\frac{\mu_{0} N i}{2 a}$
Given, $ N=50, $
$r=0.5 \,m , $
$i=2 \,A$
$\mu_{0}=4 \pi \times 10^{-7} Wb / Am$
$B=\frac{4 \pi \times 10^{-7}}{2} \times \frac{50 \times 2}{0.5}$
$=1.25 \times 10^{-4} T$
Note : Direction of magnetic field is perpendicular to the plane of the coil, that is along the axis of the coil.