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

Question

A circular coil of radius 10 cm having 100 turns carries a current of 3.2 A. The magnetic field at the center of the coil is (A) 2.01 × 10-3 T (B) 5.64 × 10-3 T (C) 2.64 × 10-4 T (D) 5.64 × 10-4 T

Tardigrade Admin · Accepted Answer

As B=(μ0 NI/2R), Here N = 100, I=3.2 A, R= 10 cm = 10×10-2 m ∴ B=(4π×10-7×100×3.2/2×0.1) =2.01×10-3 T

Q. $A$ circular coil of radius $10 c m$ having $100$ turns carries a current of $3.2 A$ . The magnetic field at the center of the coil is

$2.01 \times 1 0^{- 3} T$

$5.64 \times 1 0^{- 3} T$

$2.64 \times 1 0^{- 4} T$

$5.64 \times 1 0^{- 4} T$

As $B = \frac{μ _{0} N I}{2 R}$ ,
Here $N = 100$ ,
$I = 3.2 A$ ,
$R = 10 c m = 10 \times 1 0^{- 2} m$
$∴ B = \frac{4 π \times 1 0 ^{- 7} \times 100 \times 3.2}{2 \times 0.1}$
$= 2.01 \times 1 0^{- 3} T$

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

2.01×10−3T

5.64×10−3T

2.64×10−4T

5.64×10−4T