Question

A circular coil of radius 10 cm having 100 turns carries a current of 3.2 A. The magnetic field at the centre of the coil is:

• A. 2.01 x ${10}^{-3}T$
• B. 5.64 x ${10}^{-3}T$
• C. 2.64 x ${10}^{-4}T$
• D. 5.64 x ${10}^{-4}T$

Answer 1

Answer: (A)

2.01 x ${10}^{-3}T$

Here

$N=100,I=3.2A,R=10cm=10\times l{0}^{-2}m$

As $B$ =$\frac{{\mu }_{0}NI}{2R}$,

$\therefore B=\frac{4\pi \times {10}^{-7}\times 100\times 3.2}{2\times 0.1}=2.01\times {10}^{-3}$T