Question

A closely wound, circular coil with a diameter of 4.50 cm has 630 turns and carries a current of 0.480 A .

a)What is the magnitude of the magnetic field at the center of the coil?

b)What is the magnitude of the magnetic field at a point on the axis of the coil a distance of 7.00 cm from its center?

Answer 1

here,

a)

The magnetic field at the center is:
B = u0 * I * N/ 2R

B = 1.26 * 10^-6 * 0.48 * 630 / (2 * 1/2* 0.045)

B = 8.48 * 10^-3 Tesla

b)

we have dB = u0 *I*dL*sin(theta) / (4*pi*R^2)

the distance from the center to the point of interest is 7 cm.

The radius of the loop is 2.25 cm.

The distance from the coil segment to this point is (7^2 + 2.25^2)^(0.5) = 7.35 cm

And sin(theta) = 1 (since r and L are perpendicular)
However, the only magnetic field contribution from the loop points along the axis. So you'd need to multiply the magnetic field by cos(theta) = 2.25 / 7.35 = 0.306

Now R = 7 cm
So B = u0 *I*dL* 0.306 / (4*pi*R^2) * Circumference

B = 1.26 * 10^-6 * 0.48 * 630 * 0.306 / (2 * 0.0225) = 2.58 * 10^-3 T