Question

A circular loop of radius 13 cm carries a current of 16 A. A flat coil of radius 0.63 cm, having 48 turns and a current of 1.5 A, is concentric with the loop. The plane of the loop is perpendicular to the plane of the coil. Assume the loop's magnetic field is uniform across the coil. What is the magnitude of (a) the magnetic field produced by the loop at its center and (b) the torque on the coil due to the loop?

Answer 1

Answer:

a) Bt = 7.73 * 10^-5 T

b) T = 6.94 * 10^-7 N*m

Explanation:

Step 1: Data given

Circumar loop Radius = 13 cm

Current = 16 A

Flat coil radius = 0.63 cm

48 turns

Current = 1.5 A

a) What is the magnitude of (a) the magnetic field produced by the loop at its center

Let's assume a loop concentric with a coil, the plane of the coil is perpendicular to the plane of the loop. The magnetic field due to the loop at the center of the loop can be given by:

Bt = µ0It / 2Rt

In this case we'll get:

Bt = ((4π * 10^-7 T*m/A)(16A)) /(2*0.13m)

Bt = 7.73 * 10^-5 T

b) What is the magnitude of the torque on the coil due to the loop?

The torque magnitude excreting on the coil due to the magnetic field of the loop is given by:

T = µcBtsin(∅)

with µc = the magnetic dipole moment of the coil

with ∅ = the angle between the magnetic dipole moment and the magnetic field. The magnetic dipole moment is given by:

µc = N*Ic*A

⇒ with N = the number of turns in the coil

⇒ with A =  πRc² = the area of the coil

µc =π*N*Ic*Rc²

T= π*N*Ic*Rc²*Bt(sin∅)

In this situation we'll have:

T= π*48*1.5A* (0.63 *10^-2m)²*(7.73 * 10^-5 T)*sin(90)

T = 6.94 * 10^-7 N*m