Question

A coil of radius 10cm and having 20turns carries a current of 12A in clockwise direction when seen from east. The coil is in north south plane find the magnetic field at the center of the coil

Answer 1

Answer:

Explanation:

For coil

Current in the coil, Ix = 16A

Radius of the coil, Rx = 16cm = 0.16m

No. of turns of Coil, N �x = 20

So magnetic field due to coil X, Bx =

= 4π × 10-4 T

= 1.25

10-3 T (Towards East)

Answer 2

Given:

• Radius = 10 cm = 0.1 m
• Number of turns = 20
• Current = 12 A

To Find:

• The magnetic field at the center of the coil.

Solution:

The formula to find the magnetic field at the center of the coil is given by,

B = μnI/2r → {equation 1}

Where "B" is the magnetic field, "n" is the number of turns of the coil, "r" is the radius, "μ" is the permeability in free space (4π×$10^{-7}$ H/m), "I" is the current.

On substituting the given values in equation 1 we get,

B = (4π×$10^{-7}$  × 12× 20)/(2×0.1)

B = (3014.4×$10^{-7}$)/(0.2)

B = 15072 ×$10^{-7}$ T

∴ The magnetic field at the center of the coil =  15072 ×$10^{-7}$ T