Chapter 32, Problem 74PQ

A Figure P32.74 shows an N-turn rectangular coil of length a and width b entering a region of uniform magnetic field of magnitude B_out directed out of the page. The velocity of the coil is constant and is upward in the figure. The total resistance of the coil is R. What are the magnitude and direction of the magnetic force on the coil a. when only a portion of the coil has entered the region with the field, b. when the coil is completely embedded in the field, and c. as the coil begins to exit the region with the field?

To determine

The magnitude and direction of the magnetic force on the coil when a portion of the coil enters the region of the field.

Answer to Problem 74PQ

The magnitude of the magnetic force on the coil when a portion of the coil enters the region of the field is $\frac{N{B}^2{b}^{2}v}{R}$, and the direction is downwards along the plane of the paper.

Explanation of Solution

Faraday’s law states that an emf is induced in a coil when the magnetic flux linked with the coil changes.

The direction of the induced emf is given by Lenz law. Lenz law states that the current induced in a circuit due to change or a motion in the magnetic field, opposes the change in flux and exerts a mechanical force opposing the motion.

A current carrying conductor experiences a force in a magnetic field. Thus, the conductor experiences a force due to the current induced in it.

A coil ABCD of $N$ turns enters a magnetic field. The length of coil is $a$, breadth of the coil is $b$ and strength of the magnetic field is $B_{\text{out}}$. The magnetic field is pointing outwards that is, perpendicular to the plane of the paper.

The coil moves with a constant velocity $\overrightarrow{v}$ into the magnetic field as shown in the Figure-(1).

Figure-(1)

Write the expression for the induced emf entering the magnetic field.

    $\epsilon =N{B}_{\text{out}}bv$                                                                                                   (I)

Write the expression for magnitude of the induced current in the coil.

    $I=\frac{\epsilon }{R}$

Here, $R$ is the resistance of the coil.

Substitute equation (I) in the above equation to find $I$.

    $I=\frac{N{B}_{\text{out}}bv}{R}$                                                                                            (II)

The flux entering the coil increases, as the coil enters the magnetic field. According to Lenz law, the current in the segment which enters the field, flows towards the right, from point A to point B.

Write the expression for force experienced by the segment due to current flowing in the coil.

    $\overrightarrow{F}=I\left(\overrightarrow{b}\times {\overrightarrow{B}}_{\text{out}}\right)$

According to the right hand screw rule, the force acts perpendicular to both the direction of current and the magnetic field. Therefore, the force acts downwards.

Write the expression for magnitude of the force.

    $\begin{array}{c}F=Ib{B}_{\text{out}}\sin \theta \\ =Ib{B}_{\text{out}}\sin 90°\\ =Ib{B}_{\text{out}}\end{array}$

$\theta =90°$ as $\overrightarrow{b},{\overrightarrow{B}}_{\text{out}}$ are perpendicular to each other.

Substitute equation (II) in the above equation to find $F$.

    $F=\frac{N{\left(B_{\text{out}}\right)}^2{b}^{2}v}{R}$

Conclusion:

Therefore, the magnitude of the magnetic force on the coil, when a portion of the coil enters the region of the field is $\frac{N{\left(B_{\text{out}}\right)}^2{b}^{2}v}{R}$ and the direction is downwards along the plane of the paper.

To determine

The magnitude and direction of the magnetic force on the coil when the coil is completely embedded in the field.

Answer to Problem 74PQ

The magnitude of the force on the coil when it is completely embedded in the magnetic field is $0$.

Explanation of Solution

As shown in Figure-(2), the coil is completely embedded in the field.

Figure-(2)

According to Faraday’s law of electromagnetic induction, the emf induced in the coil is $0$, when there is no variation in the flux linked with the coil.

There is no current flow in the coil, when the induced emf is $0$. There is no occurrence of force in the coil while moving in the magnetic field, since the coil carries no current.

Conclusion:

Therefore, the magnetic force on the coil when it is completely embedded in the magnetic field is $0$.

To determine

The magnitude and direction of the magnetic force on the coil as the coil begins to exit the region of the field.

Answer to Problem 74PQ

The magnitude of the magnetic force on the coil, when a portion of the coil exits the region of the field is $\frac{N{\left(B_{\text{out}}\right)}^2{b}^{2}v}{R}$ and the direction is downwards along the plane of the paper.

Explanation of Solution

The outward flux through the coil decreases, as the coil begins to exit the field. The current flows in the counter clockwise direction in the coil. The current flows from point D to point C as shown in figure-(3).

Figure-(3)

The magnitude of the current in the coil remains the same as it is when the coil enters the field, as the coil moves with the same constant speed. The direction of the current flowing in the coil, when it exits the field, is opposite to the direction of the current when the coil enters the field.

Write the expression for the current in the coil when it exits the magnetic field.

    $I=\frac{N{B}_{\text{out}}bv}{R}$

Write the expression for magnitude of the force experienced by the coil.

    $F=\frac{N{\left(B_{\text{out}}\right)}^2{b}^{2}v}{R}$

The magnitude of the force experienced by the coil, when it exits the field is same as, when it enters the field.

The force acts downwards along the plane of the paper.

Conclusion:

Therefore, the magnitude of the magnetic force on the coil, when a portion of the coil exits the region of the field is $\frac{N{\left(B_{\text{out}}\right)}^2{b}^{2}v}{R}$ and the direction is downwards along the plane of the paper.