Question

A rectangular loop consists of 130 closely wrapped turns and has dimensions 0.400 m by 0.300 m. The loop is hinged along the y-axis, and the plane of the coil makes an angle of 30.0° with the x-axis (see figure below).

What is the magnitude of the torque exerted on the loop by a uniform magnetic field of 0.792 T directed along the x-axis when the current in the windings has a value of 1.20 A in the direction shown?
__________N · m

What is the expected direction of rotation of the loop?

A. clockwise as viewed from above.

B. counterclockwise as viewed from the above.

C.  The magnitude is zero.

Answer 1

Concepts and reason

The concepts used to solve this problem are torque and right hand rule.

First, find the magnitude of torque using the relation between the area of the rectangular loop and the angle between the magnetic moment and magnetic field.

Finally, find the expected direction using the right hand rule.

Fundamentals

A current carrying rectangular coil experiences a force in the magnetic field of each side as the current transferring conductor undergoes a force when kept in a magnetic field.

The change in an object’s angular velocity is defined as torque.

The expression for the magnitude of torque having $N$ turns in a rectangular coil is as follows:

$\tau = NBIA\sin \theta$

Here, $N$ is the number of turns, $B$ is the magnetic field, $I$ is the current, $A$ is the area, and $\theta$ is the angle between the magnetic moment and magnetic field.

Right hand rule states, “if the thumb, the middle finger and the index finger is kept such that all three are mutually perpendicular to each other, then the thumb shows the direction of force of the loop, the index finger shows the direction of the current or velocity, and the middle finger shows the direction of the magnetic field”.

(1)

The expression for the magnitude of torque having $N$ turns in a rectangular coil is as follows:

$\tau = NBIA\sin \theta$

The area of the rectangular loop is as follows:

$A = \left( l \right)\left( b \right)$

Substitute $0.30\,{\rm{m}}$ for $l$ , and $0.40\,{\rm{m}}$ for $b$ .

$\begin{array}{c}\\A = \left( {0.4\,{\rm{m}}} \right)\left( {0.3\,{\rm{m}}} \right)\\\\ = 0.12\,{{\rm{m}}^2}\\\end{array}$

The angle made by the plane of coil is $30^\circ$ .

The angle between the magnetic moment and magnetic field in a rectangular loop is as follows:

$\begin{array}{c}\\\theta = 90^\circ - 30^\circ \\\\ = 60^\circ \\\end{array}$

Substitute $0.12\,{{\rm{m}}^2}$ for $A$ , $60^\circ$ for $\theta$ , $130$ for $N$ , $0.792\,{\rm{T}}$ for $B$ , and $1.20\,{\rm{A}}$ for $I$ in the torque expression.

$\begin{array}{c}\\\tau = \left( {130} \right)\left( {0.792\,{\rm{T}}} \right)\left( {1.20\,{\rm{A}}} \right)\left( {0.12\,{{\rm{m}}^2}} \right)\sin 60^\circ \\\\ = 12.8\,{\rm{N}} \cdot {\rm{m}}\\\end{array}$

(2)

The wrong options are as follows:

• B. Counterclockwise when viewed from the above.

• C. The magnitude is zero.

According to right hand rule, the current points downward to the non-hinged side of the rectangular loop.

The magnetic field points in the positive $x$ direction of the loop.

Also, the current will point in the $z$ direction of the loop.

When viewed from the top loop, it will rotate in the clockwise direction.

Hence the correct option is as follows:

• A. Clockwise as viewed from above

Ans: Part 1

Thus, the magnitude of the torque exerted on the loop is $12.8\,N \cdot m$ .

Part 2

Clockwise as viewed from the above is the direction expected for rotation of loop.