Question

A copper coil of $100$ turns, radius $8.0 \times 10^{-2} \,m$ carries a current of $0.40 \,A$. Magnitude of magnetic field at centre of coil is

• A. $B=3.1 \times 10^{-3}\, T$
• B. $B=3.1 \times 10^{-2} \,T$
• C. $B=3.1 \times 10^{-4}\, T$
• D. $B=3.1 \times 10^{-7} \,T$

Answer 1

Answer: (C)

Magnetic field at the centre of the loop,
$B=\frac{\mu_{0} N I}{2 r}=\frac{4 \pi \times 10^{-7} \times 100 \times 0.4}{2 \times 8.0 \times 10^{-2}}$
$=3.1 \times 10^{-4} T$