Two circular loops are parallel, coaxial, and almost in contact, 1.25 mm apart (

Question

Two circular loops are parallel, coaxial, and almost in contact, 1.25 mm apart (see figure below).

Each loop is 11.9 cm in radius. The top loop carries a clockwise current of I = 126 A. The bottom loop carries a counterclockwise current of I = 126 A.

a) Calculate the magnetic force that the bottom loop exerts on the top loop.

b) The upper loop has a mass of 0.0196 kg. Calculate its acceleration, assuming that the only forces acting on it are the force in part (a) and its weight.

Two circular loops are parallel, coaxial, and almost in contact, 1.25 mm apart (see figure below). Each loop is 11.9 cm in radius. The top loop carries a clockwise current of I = 126 A. The bottom loop carries a counterclockwise current of I = 126 A. a) Calculate the magnetic force that the bottom loop exerts on the top loop. b) The upper loop has a mass of 0.0196 kg. Calculate its acceleration, assuming that the only forces acting on it are the force in part (a) and its weight.

Explanation / Answer

as these loops are parallel and very close

Force between the loops = u0*i1*i2*l/(2*pi*d)

F = 4*pi*10^-7 * 126 * 126 *2*pi*.119 /(2*pi*0.00125)

F =1.899 N

the magnetic force on the upper loop is 1.899 N upwards

b)

Now ,

Fnet = Fm - mg

Fnet = 1.899 - 0.0196 * 9.8

Fnet = 1.707 N

Now , using second law of motion

a = Fnet/m

a = 1.707/.0196

a = 87.1 m/s^2

the acceleration of the upper loop is 87.1 m/s^2

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