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You have a 7-m-long copper wire. You want to make an N-turn current loop that generates a 6.582 mT magnetic field at the center when the current is 4.092 A. You must use the entire wire. What will be the diameter of your coil?

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5 votes

Answer:

0.042 m

Step-by-step explanation:

L = length of the copper wire = 7 m

N = Number of turns in the current loop

B = magnetic field at the center = 0.006582 T

i = current flowing through the wire = 4.092 A

r = radius of the coil

length of the copper wire is given as

L = 2πrN

7 = 2πrN

rN = 1.115


N = (1.115)/(r) eq-1

Magnetic field at the center of the coil is given


B = (\mu _(o))/(4\pi ) \left ( (2\pi Ni)/(r) \right )


0.006582 = (10^(-7)) \left ( (2(3.14)(1.115)(4.092))/(r^(2)) \right )

r = 0.021 m

diameter is given as

d = 2r

d = 2 (0.021)

d = 0.042 m

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