A rectangular coil of wire 22.0 cm by 35.0 cm and carrying a current of 1.50 A i

Question

A rectangular coil of wire 22.0 cm by 35.0 cm and carrying a current of 1.50 A is oriented with the plane of its loop perpendicular to a uniform 1.60 Tmagnetic field, as shown in the figure

Part A

Calculate the net force that the magnetic field exerts on this coil.

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Part B

Calculate the torque that the magnetic field exerts on this coil.

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Part C

The coil is now rotated through a 30.0 0o angle about the axis shown, the left side coming out of the plane and the right side going into the plane. Calculate the net force that the magnetic field exerts on the coil. (Hint: In order to help visualize this three-dimensional problem, make a careful drawing of the coil as viewed along its axis of rotation.)

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Part D

Calculate the torque that the magnetic field exerts on the coil in part (C).

Explanation / Answer

(a) the area of the rctangular wire is

A = (0.22 m)*(0.35 m)
  
= 0.077 m^2

The angle theta = 0 degree, when the normal of the coil and the magnetic field parallel to each other.

Thus, the net force exerted on the coil due to the magnetic field is

F = B*I*L*sin(theta)

F = B*I*L*sin(0)

= 0

(b) The net torque exerted on the coil is

tau = B*I*A*sin(theta)

= B*I*A*sin(180)

= 0

(c) Since, the coil is still in the uniform magnetic field

the net force exerted on the coil is still zero.

F = 0

(d) The net torque exerted on the coil is

tau = B*I*A*sin(theta)

= 1.60*1.50*0.077*sin(30)

= 0.0924 N.m

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