Question

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?

Answer 1

Answer:

0.042 m

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

$N = \frac{1.115}{r}$                                             eq-1

Magnetic field at the center of the coil is given

$B = \frac{\mu _{o}}{4\pi } \left ( \frac{2\pi Ni}{r} \right )$

$0.006582 = (10^{-7}) \left ( \frac{2(3.14)(1.115)(4.092)}{r^{2}} \right )$

r = 0.021 m

diameter is given as

d = 2r

d = 2 (0.021)

d = 0.042 m