Question

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

• A. 2.4 × 10-4 T
• B. 3.5 × 10-4 T
• C. 2.79 × 10-4 T
• D. 3 × 10-4 T

Answer 1

Answer: (C)

2.79 × 10-4 T

Answer 2

Here N = 100

R = 9 cm = $9 \times 10 ^ { - 2 } \mathrm {~m}$ and

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

$= \frac { 2 \times 3.14 \times 0.4 } { 9 } \times 10 ^ { - 3 }$

$= 0.279 \times 10 ^ { - 3 } \mathrm {~T} = 2.79 \times 10 ^ { - 4 } \mathrm {~T}$