Question

You have a 1.9 m long copper wire. You want to make an N-turn current loop that generates a 1.9 mT magnetic field at the center when the current is 1.1 A . You must use the entire wire. What will be the diameter of your coil?

Answer #1

Magnetic field at the center of current carrying circular loop is given by:

B = N*u0*I/(2*R)

Given that

I = Current in loop = 1.1 A

B = Magnetic field generated by loop = 1.9 mT = 1.9*10^-3 T

L = Length of copper wire = 1.9 m

N = Number of loops, then

N*2*pi*R = 1.9 m

N = 1.9/(2*pi*R)

Using this

B = N*u0*I/(2*R)

B = 1.9*u0*I/(2*pi*R*2*R)

R^2 = 1.9*u0*I/(4*pi*B)

R = sqrt (1.9*4*pi*10^-7*1.1/(4*pi*1.9*10^-3))

R = 0.0105 m = radius of loop

**diameter of loop = 2*R = 0.0210 m = 2.1 cm**

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