http://www.austincc.edu/mmcgraw/files_2426/Review%20Sheet-2426-Exam%202.pdf 16.

Question

http://www.austincc.edu/mmcgraw/files_2426/Review%20Sheet-2426-Exam%202.pdf

16. A rectangular loop of wire (0.10 m by 0.20 m) carries a current of 5.0 A in a counterclockwise direction. The loop is oriented as shown in a uniform magnetic field of 1.5 T. The force acting on the upper 0.10-m side of the loop is A) 1.5 N B) 0.75 N C) 0.50 N D) 0.15 N E) zero

Ans: B

1-I know the answer just show the work

2- calculate the torque of the above loop?

3-what is the magnetic dipole moment associated with this loop?

Explanation / Answer

here ,

B = 1.5 T

I = 5 A

1)

force on the upper side , F= BIL

F = 1.5 * 0.1 * 5

F = 0.75 N

2)

magntic moment , u = I*A

u = 5 * (0.1 * 0.2)

u = 0.1 A.m^2

the magnetic moment of the loop is 0.1 A.m^2

3)

TOrque = u*B*sin(theta)

here , theta = 60 degree

Torque = 0.1 * 1.5 * sin(60)

Torque = 0.12 N.m

the Torque acting on the loop is 0.12 N.m

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