Question

A circular coil of wire consisting of 100 turns each of radius 9 cm carries of 0.4A. The magnitude of magnetic field at the centre of the coil is:

• A. $2.4\times {10}^{-4}T$
• B. $3.5\times {10}^{-4}T$
• C. $2.97\times {10}^{-4}T$
• D. $3\times {10}^{-4}T$

Answer 1

Answer: (C)

$2.97\times {10}^{-4}T$

Here $N=100$

$R=9cm=9\times {10}^{-2}m$ and $I=0.4A$

Now, $B=\frac{{\mu }_{0}NI}{2R}=\frac{2\pi \times {10}^{-7}\times 100\times 0.4}{9\times {10}^{-2}}$

$=\frac{2\times 3.14\times 0.4}{9}\times {10}^{-3}=0.279\times {10}^{-3}T=2.79\times {10}^{-4}T$