Question

A circular coil of wire having 200 turns, each radius 4.0 cm is placed in a horizontal plane. It carries a current of 0.40 A in the clockwise direction. Find the magnitude and direction of the magnetic field at the center of the coil.

Hint:

To find the magnetic field at the center of a coil, use the formula \( B = \frac{\mu_0 N I}{2r} \). The direction of the magnetic field can be determined using the right-hand rule.

Answer 1

The magnetic field \( B \) at the center of a circular coil with \( N \) turns, each of radius \( r \), carrying a current \( I \), is given by the formula: \[ B = \frac{\mu_0 N I}{2r} \] where:
- \( \mu_0 \) is the permeability of free space \( \mu_0 = 4\pi \times 10^{
-7} \, \text{T m/A} \),
- \( N \) is the number of turns in the coil,
- \( I \) is the current in the coil,
- \( r \) is the radius of the coil. Given:
- \( N = 200 \),
- \( I = 0.40 \, \text{A} \),
- \( r = 4.0 \, \text{cm} = 0.04 \, \text{m} \). Substitute the values into the formula: \[ B = \frac{(4\pi \times 10^{
-7}) \times 200 \times 0.40}{2 \times 0.04} \] \[ B = \frac{4\pi \times 10^{
-7} \times 80}{0.08} \] \[ B = 1.256 \times 10^{
-3} \, \text{T} \] Thus, the magnitude of the magnetic field at the center of the coil is \( 1.256 \times 10^{
-3} \, \text{T} \). For the direction, using the right
-hand rule for a coil with a clockwise current, the magnetic field will be directed downward at the center of the coil.