The magnetic field has been measured to be horizontal everyw

The magnetic field has been measured to be horizontal everywhere along a rectangular path x = 20 cm long and y = 2 cm high, shown in the figure. Along the bottom the average magnetic field B_1 = 1.6X10^-4 tesla, along the sides the average magnetic field B_2 = 1.0X10^-4 tesla, and along the top the average magnetic field B_3 = 0.5X10^-4 tesla. What can you conclude about the electric currents in the area that is surrounded by the rectangular path?

Solution

To calculate I you will need to use Ampere\'s Law for a wire.

That being: the integral of the magnetic field (B) as a dot product of dl (the distance B is across) --> B•dl.

First take B1 and multiply it by the distance B1 is across. Now multiply that by the cos of the angle between the direction of B1 and your wire (a dot product of B•dl = B*dl*cos(Ø)). Then divide this number by µ0. We know that µ0/4 = 1e-7 so µ0 must then equal 1e-7*4 = 1.2566e-6.

So for part one you have 1.6e-4T*0.20m/1.2566e-6 = 25.464.

Now we know that the I of B2 will be zero because the angle between the direction of B2 and the wire is 90° and the cos(90) = 0.

Now take B3 and do as we did to B1.

This gives you: 0.5e-4T*0.20m/1.2566e-6 = 7.957

Your final answer will be 25.464A - 7.957A = 17.507 A.

Hopes this helps!