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Chapter 19: Problem 73

Two concentric circular wire loops in the same plane each carry a current. The larger loop has a current of 8.46 A circulating clockwise and has a radius of \(6.20 \mathrm{cm} .\) The smaller loop has a radius of \(4.42 \mathrm{cm} .\) What is the current in the smaller loop if the total magnetic field at the center of the system is zero? [See Eq. \((19-16) .]\)

Short Answer

Expert verified
Answer: The current in the smaller loop is 5.07 A counterclockwise.

Step by step solution

01

Find the magnetic field due to the larger loop at the center

According to Eq. (19-16) if we have I current in a wire loop with radius R, then the magnetic field at the center of the loop is given by: \(B = \frac{\mu _{0} I}{2R}\) Where \(\mu_0 = 4\pi ×10^{-7} T.m/A\) is the permeability of free space. For the larger loop, we have I = 8.46 A (clockwise) and R = 6.20 cm: \(B_{large} = \frac{4\pi ×10^{-7} \times 8.46}{2\times0.062} \mathrm{T}\) Note that when the larger loop is clockwise current, its magnetic field will direct inside the screen (Apply right-hand rule).
02

Find the magnetic field due to the smaller loop at the center

In order to make the total magnetic field at the center zero, the magnetic field of the smaller loop should have the opposite direction to the larger loop, so the current in the smaller loop must be counterclockwise. For the smaller loop, we have R = 4.42 cm and we need to find the current I: \(B_{small} = \frac{4\pi ×10^{-7} \times I}{2\times0.0442} \mathrm{T}\) Since the magnetic field due to the smaller loop should be opposite to the magnetic field due to the larger loop and equal in magnitude: \(B_{small} = -B_{large}\)
03

Calculate the current in the smaller loop

Now we can just equate \(B_{small}\) and \(-B_{large}\), and solve for I (current in the smaller loop): \(\frac{4\pi ×10^{-7} \times I}{2\times0.0442} = -\frac{4\pi ×10^{-7} \times 8.46}{2\times0.062}\) We cancel out the \(4\pi\times10^{-7}\) on both sides and solve for I: \(I = -8.46 \times \frac{0.0442}{0.062} \mathrm{A}\) The negative sign indicates that the current flows counterclockwise, as we expected.
04

Find the current in the smaller loop

Calculate the current in the smaller loop: \(I = -8.46 \times \frac{0.0442}{0.062} = -5.07 \mathrm{A}\) So the current in the smaller loop is 5.07 A counterclockwise.

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