A rectangular coil of wire 22.0cm by 35.0cm and carrying a current of 1.60A is o

Question

A rectangular coil of wire 22.0cm by 35.0cm and carrying a current of 1.60A is oriented with the plane of its loop perpendicular to a uniform 1.60Tmagnetic field, as shown in the figure (Figure 1)

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

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

C. The coil is now rotated through a 30.00o 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.)

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

A rectangular coil of wire 22.0cm by 35.0cm and carrying a current of 1.60A is oriented with the plane of its loop perpendicular to a uniform 1.60Tmagnetic field, as shown in the figure (Figure 1) A. Calculate the net force that the magnetic field exerts on this coil. B. Calculate the torque that the magnetic field exerts on this coil. C. The coil is now rotated through a 30.00o 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.) D. Calculate the torque that the magnetic field exerts on the coil in part (C).

Explanation / Answer

Here ,

Inside a magnetic field , the net force acting on a loop is zero.

hence , the force acting on the loop is ZERO.

B)

Now ,

as the field is perpendicular to loop

torque = N*B*A*I* cos(theta)

torque = N*B*A*I* cos(90)

torque = 0 N.m

C)

Here , now , the force on this closed loop is zero.

D)

Here , theta =30

torque = N*B*A*I* cos(theta)

torque = 0.22 * 0.35 * 1.6 *1.6 cos(30)

torque = 0.171 N.m

the torque acting on the loop is 0.171 N.m

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