This question has 3 parts: Two straight parallel wires carry currents in opposit

Question

This question has 3 parts:

Two straight parallel wires carry currents in opposite directions as shown in the figure. One of the wires carries a current of I2 = 11.8 A. Point A is the midpoint between the wires. The total distance between the wires is d = 10.2 cm. Point C is 5.85 cm to the right of the wire carrying current I2. Current I1 is adjusted so that the magnetic field at C is zero. Calculate the value of the current I1.

Calculate the magnitude of the magnetic field at point A.

What is the force between two 1.23 m long segments of the wires?

Explanation / Answer

I1 = 11.8/5.85 *(10.2+5.85) = 32.37 A

distance of A from both wire = d/2 = 10.2/2 = 5.1cm
at point A
B1 = u*I1/2*pi*r (outside the plain)
B2= u*I2 / 2*pi*r ( outside the plain)


so total B = u*(I1+i2)/ 2*pi*r =
= 4*pi*10^-7 *(32.37+11.8) / 2*pi*0.051 = 17.32*10^-5 T


force between them
F/L = Mu*I1*I2/ 2*pi*r

F = 1.23 * 4*pi*10^-7 * 32.37*11.9 / 2*pi*0.102
= 0.929 *10^-3 N

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