Question

help please!

Answer 1

Given that,

Magnitude of the current = I = 1.5 A

Directed right to left.

A)from the thumb rule, when we point the thumb in the direction of the current, the curved fingers will point in the direction of magnetic field. So the current in this case is flowing from right to left, hence the field will be pointed into the page.

(Apologies, no success in uploading the picture).

B) r = 0.125 meters

We know that, the magnetic field due to a current carrying wire at a distance r is given by:

B = $\mu$o I / 2 pi r = 4 pi x 10^-7 x 1.5 / 2 pi x 0.125 = 2.4 x 10^-6 T = 2.4 $\mu$T

Hence, B = 2.4 $\mu$T

C)Thw direction of current will be counter clokwise. Because the magnetic field is into the page. And when the loop will be placed, an induced emf and hence an induced current will be generated which will be in the oppsite direction.

D)N = 325 turns ; length of side = a = 0.65 m; B = 0.0365 T/s so, t = 1 sec and R = 12.5 Ohm

We know from Faraday's law of induction that,

$\varepsilon$ = - d$\Phi$/dt = $\Phi$1 - $\Phi$2 / 1 sec

we know that, $\Phi$ = B A = 0.0365 x (0.65)² = 0.0154 Weber.

$\varepsilon$ = - ( 0 - 0.0154)/1 = 0.0154

We know that, V = IR => I = V/R

I = $\varepsilon$/R = 0.0154 / 12.5 = 0.001232 Ampere

Hence, induced current = I = 0.001232 Ampere.