Question

5) A rectangular coil of wire, 22.0 cm by 35.0 cm
and carrying a current of 1.40 A, is oriented with
the plane of its loop perpendicular to a uniform
1.50-T magnetic field, as shown in the figure.
a) Calculate the net force which the magnetic field
exerts on the coil
b) Calculate the torque which the magnetic field
exerts on the coil
c) The coil is rotated through a 30.0degrees angle about the
axis shown, the left side coming out of the plane of the figure and the right
side going into the plane. Calculate the net force which the magnetic field
now exerts on the coil. (Hint: In order to help visualize this 3-dimensional
problem, make a careful drawing of the coil when viewed along the
rotation axis.)
d) Calculate the torque which the magnetic field now exerts on the coil

Answer 1

Concepts and reason

The concepts required to solve this problem are force on a current carrying wire in magnetic field, torque on a current carrying loop and magnetic moment.

First use the equation of force acting on a wire due to magnetic field to solve for the net force on the rectangular coil.

Later substitute the moment in the torque equation and calculate the torque on the rectangular coil. Again, solve for the force and torque for the rotated rectangular coil.

Fundamentals

The force on a straight carrying wire in the magnetic field is,

$F = Bil\sin \alpha$

Here, $B$ is the perpendicular magnetic field, $i$ is the current, $l$ is the length of the wire in the magnetic field, and $\alpha$ is the angle between the direction of current and magnetic field.

The magnetic moment of a current carrying loop is,

$\mu = iA$

Here, $i$ is the current, and $A$ is the area.

The area of a rectangle is,

$A = lb$

Here, $l$ is the length, and $b$ is the breadth.

The torque on the current carrying loop is,

$\tau = \mu B\sin \theta$

Here, $\mu$ is the magnetic moment, $B$ is magnetic field strength, and $\theta$ is the angle between the magnetic moment vector and magnetic field.

(a)

The magnetic field is perpendicular to all the sides of the rectangular coil. Therefore, there is non-zero force on all the sides of the rectangular coil. The direction of force is opposite on the opposite sides of the rectangular coil as direction of current is opposite but the magnitude is same as all other parameters are same for the opposite sides of a rectangle.

So, the opposite force balance out for the whole rectangular coil and thus, net force which the magnetic field exerts on the coil is zero.

(b)

Use the equation of torque in magnetic field.

Substitute $0^\circ$ for $\theta$ in the equation $\tau = \mu B\sin \theta$.

$\begin{array}{c}\\\tau = \mu B\sin 0^\circ \\\\ = 0\\\end{array}$

(c)

The magnetic field is at angle $30^\circ$ to all the sides of the rectangular coil. Therefore, there is non-zero force on all the sides of the rectangular coil. The direction of force is opposite on the opposite sides of the rectangular coil as direction of current is opposite but the magnitude is same as all other parameters are same for the opposite sides of a rectangle.

So, the opposite force balance out for the whole rectangular coil and thus, net force which the magnetic field exerts on the coil is zero.

(d)

Use the magnetic moment equation.

Substitute $lb$ for $A$ in the magnetic moment equation $\mu = iA$.

$\mu = ilb$

Substitute $ilb$ for $\mu$ in the torque equation $\tau = \mu B\sin \theta$.

$\tau = ilbB\sin \theta$

Substitute $1.40{\rm{ A}}$ for $i$, $22.0{\rm{ cm}}$ for $l$, ${\rm{35}}{\rm{.0 cm}}$ for $b$, $1.50{\rm{ T}}$ for $B$, and $30^\circ$ for $\theta$ in the above equation $\tau = ilbB\sin \theta$ and calculate the torque on the loop.

$\begin{array}{c}\\\tau = \left( {1.40{\rm{ A}}} \right)\left( {22.0{\rm{ cm}}} \right)\left( {35.0{\rm{ cm}}} \right)\left( {{\rm{1}}{\rm{.50 T}}} \right)\sin 30^\circ \\\\ = \left( {{\rm{1}}{\rm{.40 A}}} \right)\left( {22.0{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)\left( {35.0{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)\left( {{\rm{1}}{\rm{.50 T}}} \right)\left( {0.5} \right)\\\\ = 0.081{\rm{ N}} \cdot {\rm{m}}\\\end{array}$

Ans: Part a

The net force which the magnetic field exerts on coil is $0$.

Part b

The torque which magnetic field exerts on the coil is $0$.

Part c

The net force which the magnetic field exerts on coil is $0$.

Part d

The torque which magnetic field exerts on the coil is $0.081{\rm{ N}} \cdot {\rm{m}}$.