Question

A long straight wire carries a current of 10.5 A. A second wire is bent into a loop, with a radius of R = 7.5 cm. The wire loop is held at a distance d = 3.0 cm away from the long straight wire. Find the magnitude and direction of the currrent in the loop which will produce a net magnetic field of 6.7 x 10-6 T out of the page at the center of the loop.

Answer 1

Answer:

i = 3.18 A  anticlockwise

Explanation:

magnetic field due to wire at centre$= \frac{ \mu_o I}{2 \pi r}$

$= \frac{ 4*\pi *10^{-7} *10.5}{2 \pi*10^{-2}} = 2*10^{-5} T$

Current in the loop can be determined by using magnetic field formula in centre of loop

$= \frac{ 4*\pi *10^{-7} *I}{2 \pi*7.5^{-2}}$

total net magnitude is given as 6.7*10^[-6} T

SO WE HAVE

$6.7*10^{-6} = \frac{ 4*\pi *10^{-7} *I}{2 \pi*7.5^{-2}} -2*10^{-5}$

solving above equation we get

i = 3.18 A anticlockwise to have magnetic field direction out of the page