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

i = 3.18 A anticlockwise

Step-by-step explanation:

magnetic field due to wire at centre
`= ( \mu_o I)/(2 \pi r)`

`= ( 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

`= ( 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) = ( 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