Question

A tightly wound coil of 200 turns and of radius 20 cm carrying current 5 A. Magnetic field at the centre of the coil is.

• A. \( 3.14 \times 10^{-3} \, \text{T} \)
• B. \( 3.14 \times 10^{-2} \, \text{T} \)
• C. \( 6.28 \times 10^{-4} \, \text{T} \)
• D. \( 6.28 \times 10^{-3} \, \text{T} \)

Hint:

The magnetic field at the center of a current-carrying coil can be found using the formula \( B = \frac{\mu_0 N I}{2r} \), where \( N \) is the number of turns, \( I \) is the current, and \( r \) is the radius of the coil.

Answer 1

The Correct Option is A

The magnetic field at the center of a coil of \( N \) turns with radius \( r \) and current \( I \) is given by the formula: \[ B = \frac{\mu_0 N I}{2r} \] Where \( \mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A} \), \( N = 200 \), \( I = 5 \, \text{A} \), and \( r = 0.2 \, \text{m} \). Substituting the values: \[ B = \frac{4 \pi \times 10^{-7} \times 200 \times 5}{2 \times 0.2} = 3.14 \times 10^{-3} \, \text{T} \] Thus, the magnetic field at the center of the coil is \( 3.14 \times 10^{-3} \, \text{T} \).