A long, straight wire carries a current of 2 A to the left. Next to the wire is

Question

A long, straight wire carries a current of 2 A to the left. Next to the wire is a square loop with sides 1.0 m in length, as shown below. The loop carries a current of 2.5 A in the direction indicated.

(a) What is the net force exerted on the loop?
size: 4.16666 N   CORRECT
direction: away from the wire  CORRECT

(b) What is the direction of the net torque on the loop, as viewed from the left? NOTE THAT THIS IS NOT A UNIFORM FIELD, PLEASE EXPLAIN YOUR ANSWER
clockwise / torque is zero   /not enough info

Explanation / Answer

given that

Ia = 2A ( wire carries a current )

Ib = 2.5A (loop carries a current )

length of loop's sides L = 1 m

distance b/w looop and wire r = 0.2m

part(a)

we know that

magnetic force F = Ib*(L x B)

B =uo*Ia / 2*pi*r

So the wire induces a B field that point inward into the screen, and this induces a net force on the square loop only for the part of the square loop that is parallel to the long straight wire. The part of the square loop that is perpendicular to the long wire has zero net force acting on it

So for the part of the square loop that's closer to the wire, r = 0.2 m.

B1 = (4*pi * 10^-7)(2) / 2*pi*(0.2) = 2*10^(-6) T

For the part of the square loop that's further away from the wire, r = 1.2 m

B2 = (4*pi*10^-7)(2) / 2*pi*(1.2) = 3.33*10^-7 T

Now the net force acting on the square loo is

F = Ib*L*(B1 - B2)

F = (2.5)*1* (2*10^-6 - 3.33*10^-7) = 4.17* 10^-6 N

direction : away fron the wire

part(b)

we know that

torque = I (A x B)

where A is area of loop .

B is magnetic fieild.

direction of torque is A x B.

A is pependicular to plane and B is also perpendicular to plane .

so A and B are parallel to each other .

so torque is zero.

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