A single-turn wire loop produces a magnetic field of 41.2 muT at its center, and

Question

A single-turn wire loop produces a magnetic field of 41.2 muT at its center, and 5.15 nT on its axis, at 24.0 cm from the loop center. Find the loop radius. Express your answer with the appropriate units. Find the current in the loop. Express your answer with the appropriate units. In class and in the textbook, we used the Biot-Savart Law to express the magnetic field dB of a current element of length de carrying current I. In this problem, we are finding the field B of a charge q moving with speed u. The quantity I de from the standard Biot-Savart Law should be replaced with what quantity?

Explanation / Answer

Magnetic Field due to current carrying Loop at the center of the loop,
B = (uo*I)/(2*a)
41.2 * 10^-6 = (2*pi*10^-7) * (I/a)
(I/a) = 65.6 --------1

Magnetic Field due to current carrying Loop along the axis is given by,
B = uo/2 * Ia^2/(x^2 + a^2)^3/2
5.15 * 10^-9  = (2*pi*10^-7) * Ia^2/(0.24^2 + a^2)^3/2
Ia^2/(0.24^2 + a^2)^3/2 = 0.0082 -------2

Solving eq 1&2
I = 0.79 A
a = 0.012 m

Loop radius, a = 0.012 m
Current in the loop, I = 0.79 A

From Bio Savart Law,
dB = uo/4*pi (I.dl x r^) / r^2
For Charge moving with speed v,
dB =  uo/4*pi (qv x r^) / r^2

So I.dl is replaced by , q*v

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