need these three COMPLETE and final answer should be correct! 8. + Question Deta

Question

need these three COMPLETE and final answer should be correct!

8. + Question Details SerCP8 19.P.062. My Notes Ask Your T A single-turn square loop of wire 1.00 cm on a side carries a counterclockwise current of 0.300 A. The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 50 turns per centimeter and carries a counterclockwise current of 10.0 A. Find the force on each side of the loop and the torque acting on the loop. N (top side) N (bottom side) N (left side) N (right side) N-m (torque)

Explanation / Answer

8.

We know that magnetic field due to solenoid is B =?0nI

?0 = 4?*10-7H/m

n = number of turns per unit length = 50 turns/cm =5000turns/m

I = current = 10 A

Then B = ( 4?*10-7H/m)*5000turns/m*10 A =6.28*10-2T

Here current in the solenoid is in counterclockwise directionand hence magnetic field due to solenoid is out of page.

Now current in the single square loop is i = 0.3 A

Side of the square is L = 1 cm = 0.01 m

N = number of turns of the loop = 1

Then force on top side is F top= NBiL = 6.28*10-2T*0.3A*0.01 m = 18.8 *10-5N

Then force on bottom side is F bottom= N BiL = 6.28*10-2T*0.3A*0.01 m = 18.8 *10-5N

Then force on left side is F left= NBiL = 6.28*10-2T*0.3A*0.01 m = 18.8 *10-5N

Then force on right side is F right= N BiL = 6.28*10-2T*0.3A*0.01 m = 18.8 *10-5N

Then torque on the  single-turn square loop ofwire ? = NIABsin?

sin? = sin0 = 0

? = angle between normal to the plane of the loopand magnetic field

Then torque acting on the loop is ? = 0 Nm

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