Question

A rectangular coil of wire, 22.0 cm by 35.0 cm and carrying a current of 1.95 A, is oriented with the plane of its loop perpendicular to a uniform 1.50-T magnetic field (Fig. E27.42). (a) Calculate the net force and torque that the magnetic field exerts on the coil. (b) The coil is rotated through a 30.0° angle about the axis shown, with the left side coming out of the plane of the figure and the right side going into the plane. Calculate the net force and torque that the magnetic field now exerts on the coil. (Hint: To visualize this three-dimensional problem, make a careful drawing of the coil as viewed along the rotation axis.)

Short answer

(a) the net force and torque that the magnetic field exerts on the coil F=0

(b) the net force and torque that the magnetic field now exerts on the coil F=0 .

Step-by-step solution

Definition of torque

It measures of force that can cause an object to rotate about an axis.

a=35.0cm

b=22.0cm

I=1.95A

B=1.50T

Step 2:

(a)for the net force and torque that the magnetic field exerts on the coil.

Force acting both right and left side has same magnitude but in opposite direction. also, force acting on upper and lower side have same magnitude in opposite direction. Force produces no torque and net force acting is also zero. Force on each side points at same direction of position vector.$\mathrm{\varphi }=0°,\sin \left(\mathrm{\varphi }\right)=0$ thus,

F=0 and$\mathrm{\tau }=0$

Thus, the net force and torque that the magnetic field exerts on the coil F=0

Step 3: 

(b)for the net force and torque that the magnetic field now exerts on the coil.

The coil is rotated at $\mathrm{\varphi }=30.0°$about axis. The force in upper and lower side has same magnitude in opposite direction which cancel each other.so net force is zero but as torque is there because angle between force on each side position vector is ,

$\mathrm{\varphi }=30.0°$

$\begin{array}{r}\mathrm{T}=\mathrm{IBAsin}\left(\mathrm{\varphi }\right)\\ \mathrm{A}=\mathrm{ab},\\ \mathrm{T}=(1.95\mathrm{A})(1.50\mathrm{T})(0.220\mathrm{m})(0.350\mathrm{m})\sin (30.0)\\ \mathrm{T}=0.113\mathrm{N}\times \mathrm{m}\end{array}$

Thus, the net force and torque that the magnetic field now exerts on the coil F=0 .

$\mathrm{\tau }=0.113\mathrm{N}.\mathrm{m}$