Question

A tightly wound 90 turn coil of radius 15 cm has a magnetic field of 4 × 10^-4 T at its center. The current flowing through it is
1. 1.06 A
2. 2.44 A
3. 3.44 A
4. 4.44 A

Answer 1

Correct Answer - Option 1 : 1.06 A

Concept:

The magnetic field at point O due to the circular coil is given by:

B = \(μ_0 I \over{2R}\)

where,

B is the magnetic field at the center, μ0 permeability of the medium I is the current in the circular loop, and R is the radius of the circular coil.

Calculation:

Here, N = 90

R = 15 cm = 15 × 10^-2 m, B = 4 × 10^-4 T

∵ \(B = \frac{{{\mu _0}NI}}{{2R}}\)

∴ \(I = \frac{{2RB}}{{N{\mu _0}}}\)

\(= \frac{{2\; \times\; 15\; \times\; {{10}^{ - 2}\;} \times \;4\; \times\; {{10}^{ - 4}}}}{{4\pi \; \times\; {{10}^{ - 7}}\; \times\; 90}}\)

I = 1.06 A