Question

A closely wound, circular coil with radius 2.80 cmhas 800 turns of wire. "Closely Wound" means all of the wire loops can be approximated to be the same radius and at the same position.

A) What must the current in the coil be if the magnetic field at the center of the coil is 0.0790 T?

I=

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?

Express your answer to three significant figures

x=

Answer 1

Magnetic field at the center of circular loop is given by:

B = N*u0*i/2r

i = 2*r*B/(u0*N)

Using given values:

i = 2*2.8*10^-2*0.0790/(4*pi*10^-7*800)

i = 4.4 Amp

Part B.

Now magnetic field along the axis of loop is given by:

B = u0*i*N*r^2/(2*(r^2 + x^2)^(3/2))

Now here B = 0.0790/2 = 0.0395 T

(r^2 + x^2)^(3/2) = u0*i*N*r^2/(2B)

(r^2 + x^2) = [u0*i*N*r^2/(2B)]^(2/3)

x^2 = [u0*i*N*r^2/(2B)]^(2/3) - r^2

Using known values:

x = sqrt [[u0*i*N*r^2/(2B)]^(2/3) - r^2]

x = sqrt [[4*pi*10^-7*4.4*800*0.028^2/(2*0.0395)]^(2/3) - 0.028^2]

x = 0.0215 m = 2.15 cm

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