A copper coil of 100 turns, radius 8.0 × 10-2 m carries a current of 0.40 A. Magnitude of magnetic field at centre of coil is

Question

A copper coil of 100 turns, radius 8.0 × 10-2 m carries a current of 0.40 A. Magnitude of magnetic field at centre of coil is (A) B=3.1 × 10-3 T (B) B=3.1 × 10-2 T (C) B=3.1 × 10-4 T (D) B=3.1 × 10-7 T

Tardigrade Admin · Accepted Answer

Magnetic field at the centre of the loop, B=(μ0 N I/2 r)=(4 π × 10-7 × 100 × 0.4/2 × 8.0 × 10-2) =3.1 × 10-4 T

Q. A copper coil of $100$ turns, radius $8.0 \times 1 0^{- 2} m$ carries a current of $0.40 A$ . Magnitude of magnetic field at centre of coil is

$B = 3.1 \times 1 0^{- 3} T$

$B = 3.1 \times 1 0^{- 2} T$

$B = 3.1 \times 1 0^{- 4} T$

$B = 3.1 \times 1 0^{- 7} T$

Magnetic field at the centre of the loop,
$B = \frac{μ _{0} N I}{2 r} = \frac{4 π \times 1 0 ^{- 7} \times 100 \times 0.4}{2 \times 8.0 \times 1 0 ^{- 2}}$
$= 3.1 \times 1 0^{- 4} T$

Q. A copper coil of 100 turns, radius 8.0×10−2m carries a current of 0.40A. Magnitude of magnetic field at centre of coil is

B=3.1×10−3T

B=3.1×10−2T

B=3.1×10−4T

B=3.1×10−7T