The magnetic field has been measured to be horizontal everywhere along a rectang

Question

The magnetic field has been measured to be horizontal everywhere along a rectangular path x = 16 cm long and y = 4 cm high, shown in the figure. Along the bottom the average magnetic field B1 = 1.7 x 10-4 tesla, along the sides the average magnetic field B2 = 0.9 × 10-4 tesla, and along the top the average magnetic field B3 = 0.6 x 10-4 tesla. what can you conclude about the electric currents in the area that is surrounded by the rectangular path? 1 10.186 X A B3 B1 Additional Materials Section 21.6

Explanation / Answer

Here, use Ampere's Law to find the current in a wire.

According to this law, 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.

Now, we know that µ0/4? = 1 x 10^-7

so µ0 = 1.0 x 10^-7 x 4? = 1.2566 x 10^-6.

Therefore for the bottom portion , we have the current -

I1 = (1.70 x 10^-4 x 0.16m) / (1.2566 x 10^-6) = 21.65 A

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.

So, I2 = 0

Now take B3 and do as we did to B1.

This gives you: I3 = (0.60 x 10^-4T x 0.16m) / (1.2566 x 10^-6) = 7.64 A

Therefore, the resulting current I = I1 - I3 = 21.65 - 7.64 = 14.01 A

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