## A conventional current of 8 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius 0.078 m. Ano

Question

A conventional current of 8 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius 0.078 m. Another circular loop of wire lies in the same plane, with its center at the origin and with radius 0.03 m. How much conventional current must run counterclockwise in this smaller loop in order for the magnetic field at the origin to be zero

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1 year 2021-08-20T18:12:35+00:00 1 Answers 0 views 0

I2 = 3.076 A

Explanation:

In order to calculate the current in the second loop, you take into account that the magnitude of the magnetic field at the center of the ring is given by the following formula:

$$B=\frac{\mu_oI}{2R}$$        (1)

I: current in the wire

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

In the case of the two wires with opposite currents and different radius, but in the same plane, you have that the magnitude of the magnetic field at the center of the rings is:

$$B_T=\frac{\mu_oI_1}{2R_1}-\frac{\mu_oI_2}{2R_2}$$         (2)

I1: current of the first ring = 8A

R1: radius of the first ring = 0.078m

I2: current of the second ring = ?

R2: radius of the first second = 0.03m

To find the values of the current of the second ring, which makes the magnitude of the magnetic field equal to zero, you solve the equation (2) for I2:

$$\frac{\mu_oI_2}{2R_2}=\frac{\mu_oI_1}{2R_1}\\\\I_2=I_1\frac{R_2}{R_1}=(8A)\frac{0.03m}{0.078m}=3.076A$$

The current of the second ring is 3.076A and makes that the magntiude of the total magnetic field generated for both rings is equal to zero.