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

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

Tardigrade Admin · Accepted Answer

Here N= 100 R=9 cm =9×10-2 m and I=0.4 A Now, B=(μ0NI/2R)=(2π×10-7×100×0.4/9×10-2) =(2×3.14×0.4/9)×10-3 =0.279×10-3 T=2.79×10-4 T

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

$2.4 \times 1 0^{- 4} T$

$3.5 \times 1 0^{- 4} T$

$2.79 \times 1 0^{- 4} T$

$3 \times 1 0^{- 4} T$

Here $N = 100$
$R = 9 c m = 9 \times 1 0^{- 2} m$ and $I = 0.4 A$
Now, $B = \frac{μ _{0} N I}{2 R} = \frac{2 π \times 1 0 ^{- 7} \times 100 \times 0.4}{9 \times 1 0 ^{- 2}}$
$= \frac{2 \times 3.14 \times 0.4}{9} \times 1 0^{- 3} = 0.279 \times 1 0^{- 3}$
$T = 2.79 \times 1 0^{- 4} T$

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

2.4×10−4T

3.5×10−4T

2.79×10−4T

3×10−4T