Question

A closely wound, circular coil with radius 2.90 cmcm has 770 turns. (a) What must the current in the coil be if the magnetic field at the center of the coil is 0.0770 T? (b) At what distance x from the center of the coil, on the axis of the coil, is the magnetic field half its value at the center?

Answer 1

Answer:

(a) Current flowing through the coil is 4.61 A

(b) Distance from the center of coil is 0.035 m

Explanation:

Given :

Radius of circular coil, r = 2.90 cm = 2.90 x 10⁻² m

Number of turns, N = 770

(a) Magnetic field at the center of coil, B = 0.077 T

Consider I be the current flowing through the circular coil.

The magnetic field at the center of the circular coil is given by the relation:

$B=\frac{\mu_{0} NI}{2R}$

Substitute the suitable values in the above equation.

$0.077=\frac{4\pi\times10^{-7} \times770\times I}{2\times2.9\times10^{-2} }$

I = 4.61 A

(b) Consider x be the distance from the coil where the magnetic field due to circular coil is half of the magnetic field at its center, i.e.,

B₁ = B/2    ...(2)

Magnetic field on the axis of the coil is given by the relation:

$B_{1}=\frac{\mu_{0} NIR^{2} }{(x^{2} +R^{2})^{3/2} }$

Put equation (2) in the above equation.

$\frac{B}{2} =\frac{\mu_{0} NIR^{2} }{(x^{2} +R^{2})^{3/2} }$

Substitute the suitable values in the above equation.

$\frac{0.077}{2} =\frac{4\pi\times10^{-7} \times770\times4.61\times(2.90\times10^{-2}) ^{2} }{(x^{2} +(2.90\times10^{-2})^2)^{3/2} }$

x = 0.035 m