1. Two large, flat current-carrying sheets are placed parallel to each other, on

Question

1. Two large, flat current-carrying sheets are placed parallel to each other, one sheet above the other. The upper sheet carries a current 2.0 A per unit of width to the left, and the lower sheet carries a current 7.5 A per unit of width to the right.

Calculate the magnitude of the magnetic field between the sheets.

Calculate the magnitude of the magnetic field above the upper sheet.

Calculate the magnitude of the magnetic field below the lower sheet.

2. You need to use a long solenoid to produce a magnetic field of magnitude 0.055 T .

If the maximum current you are able to run through the windings is 20 A, what is the minimum number of windings per meter the solenoid must have?

Explanation / Answer

Calculate the magnitude of the magnetic field between the sheets.

magnetic field due to upper wire B1 = -(1/2)*uo*K1

magnetic field due to lower wire B2 = -(1/2)*uo*K2

B = B1 + B2 = -(1/2)*uo*k1 - (1/2)*uo*k2

B = -((1/2)*4*pi*10^-7*2) -( (1/2)*4*pi*10^-7*7.5)

B = 5.96*10^-6 T

_________________

Calculate the magnitude of the magnetic field above the upper sheet

magnetic field due to upper wire B1 = +(1/2)*uo*K1

magnetic field due to lower wire B2 = -(1/2)*uo*K2

B = B1 + B2 = -(1/2)*uo*k1 - (1/2)*uo*k2

B = ((1/2)*4*pi*10^-7*2) -( (1/2)*4*pi*10^-7*7.5)

B = -3.456*10^-6 T

_________________

Calculate the magnitude of the magnetic field below the lower sheet

magnetic field due to upper wire B1 = -(1/2)*uo*K1

magnetic field due to lower wire B2 = +(1/2)*uo*K2

B = B1 + B2 = -(1/2)*uo*k1 - (1/2)*uo*k2

B = -((1/2)*4*pi*10^-7*2) + ( (1/2)*4*pi*10^-7*7.5)

B = 3.456*10^-6 T

++++++++++++++++++

(2)

magnetic field due to a solenoid B = uo*n*I

0.055 = 4*pi*10^-7*n*20

n = 2188 turns/m <<<<_----------answeer

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