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1. The two wires shown in the figure carry currents of I = 5.33 A in opposite di

ID: 1284384 • Letter: 1

Question

1. The two wires shown in the figure carry currents of I = 5.33 A in opposite directions and are separated by a distance of d0 = 9.63 cm. Calculate the net magnetic field at a point midway between the wires. Use the direction out of the page as the positive direction and into the page as the negative direction in your answer.

2. Calculate the net magnetic field at point P1 - that is 8.02 cm to the right of the wire on the right.

3. Calculate the net magnetic field at point P2 - that is 18.9 cm to the left of the wire on the left.

1. The two wires shown in the figure carry currents of I = 5.33 A in opposite directions and are separated by a distance of d0 = 9.63 cm. Calculate the net magnetic field at a point midway between the wires. Use the direction out of the page as the positive direction and into the page as the negative direction in your answer. 2. Calculate the net magnetic field at point P1 - that is 8.02 cm to the right of the wire on the right. 3. Calculate the net magnetic field at point P2 - that is 18.9 cm to the left of the wire on the left.

Explanation / Answer

I'll solve this assuming the left wire is up and the right wire is down. If this is not the case basically the magnitude will be the same but the direction will be opposite

In my case both fields will be into the page and will have the same magnitude

So B = 2*(?o*I/2?r) = (but ?o/2? = 2.0X10^-7)

so B = 2*(2.0x10^-7*5.33/(0.0963/2) = 4.42x10^-5T (IN)

b) Here the left wire creates a field into the page and the right wire is out

so the net field will be IN

B = ?o/2?*(I1/0.0802 - I2/(0.0963 + 0.0802)) = 2.0x10^-7*(5.33/0.0802 - 5.33/(0.0963 + 0.0802) =

= 7.25x10^-6 T IN

c) Now the net field will be OUT

B = ?o/2?*(I1/0.189 - I2/(0.0963 + 0.189)) = 2.0x10^-7*(5.33/0.189 - 5.33/(0.0963 + 0.189) =

= 1.903x10^-6 T OUT