Question

A coil has \({N_{a}}\) turns enclosing an area of A. In a physics laboratory experiment, the coilis rotated during the time interval \({\Delta t}\) from a position in which theplane of each turn is perpendicular to Earth's magnetic field toone in which the plane of each turn is parallel to the field. Themagnitude of Earth's magnetic field at the lab location is B.

a.)What is the total magnitude \({\Phi_\text { initial }}\) of the magneticflux through the coil before it is rotated?

b.)What is the magnitude \(\Phi_{\text {final }}\) of the total magneticflux through the coil after it is rotated?

C.)What is the magnitude of the average emf induced in the coil?

Answer 1

Concepts and reason

The concepts used to solve this problem are magnetic flux and faraday's law of induction. Use the relationship between magnetic field, area of the coil, and the angle between the magnetic field and the area of the coil to calculate the magnetic flux through the coil before and after it is rotated. Finally use faraday's law of induction to calculate the average induced emf in the coil.

Fundamentals

The magnetic flux is a measure of the number of magnetic field lines passing through an area (such as a loop of wire). Expression for the magnetic flux is, \(\phi=B A \cos \theta\)

Here, \(\Phi\) is the magnetic flux, \(B\) is the magnetic field, \(A\) is the area of the coil, and \(\theta\) is the angle between the magnetic field vector and the area vector. Lenz's law states that the induced emf of the coil is equal to the negative of the rate of change of magnetic flux times the number of turns in the coil. Expression for the magnitude of average emf induced in the coil during rotation is, \(\varepsilon=N_{a} \frac{(\Delta \phi)}{\Delta t}\)

Here, \(\varepsilon\) is the emf of the coil, \(N_{a}\) is the number of loops and \(\Delta \phi / \Delta t\) is the rate of change of magnetic flux.

 

(a) Expression for the initial magnetic flux is,

\(\phi_{\text {initial }}=B A \cos \theta_{i}\)

Here, \(\phi_{\text {initial is the initial magnetic flux before rotation and } \theta_{i} \text { is the initial angle between the direction of area vector }}\) and the magnetic field vector before rotation. Initially, the plane of the coil is perpendicular to the magnetic field. since the direction of the area vector is parallel to the magnetic field vector.

Substitute \(0^{\circ}\) for \(\theta_{i}\)

\(\phi_{\text {initial }}=B A \cos \left(0^{\circ}\right)\)

=B A

Part a

Total magnitude \(\phi_{\text {initial of the magnetic flux through the coil before it is rotated is }}\) BA.

The initial angle before rotation is the position in which the plane of each turn is perpendicular to Earth's magnetic field. In the case that the surface is perpendicular to the magnetic field then the angle is zero. Hence the initial magnetic flux of the coil before its rotation is said to be maximum.

 

(b) Expression for the final magnetic flux is,

\(\phi_{\text {final }}=B A \cos \theta_{f}\)

Here, \(\phi_{\text {final is the final magnetic flux after rotation and }} \theta_{f}\) is the final angle between the direction of area vector and the magnetic field vector after rotation. Finally, the plane of the coil is parallel to the magnetic field. Since the direction of the area vector is perpendicular to the magnetic field vector.

Substitute \(90^{\circ}\) for \(\theta_{f}\)

\(\phi_{\text {final }}=B A \cos \left(90^{\circ}\right)\)

\(=0 \mathrm{~Wb}\)

Part \(b\)

Total magnitude \(\phi_{\text {linal }}\) of the magnetic flux through the coil after it is rotated is 0 Wb.

The final angle after rotation is the position in which the plane of each turn is parallel to Earth's magnetic field.

In the case that the surface is parallel to the magnetic field then the angle is \(90^{\circ} .\) Hence the final magnetic flux of the coil after its rotation is said to be minimum.

 

(c)

$$ \Delta \phi=\phi_{\text {final }}-\phi_{\text {initial }} $$

Substitute 0 for \(\phi_{\text {final }}\) and \(B A\) for \(\phi_{\text {initial } .}\)

$$ \begin{aligned} \Delta \phi &=0-B A \\ &=B A \end{aligned} $$

The sign is ignored since the magnitude is to be calculated. Expression for the average induced emf of the coil is,

\(\varepsilon=N_{a} \frac{\Delta \phi}{\Delta t}\)

Substitute BA for \(\Delta \phi\) in the above expression.

\(\varepsilon=N_{a} \frac{(0-B A)}{\Delta t}\)

\(=\frac{N_{a} B A}{\Delta t}\)

The magnitude gives only the amount of emf induced and not the direction. So, sign is ignored.

Part \(\mathrm{c}\)

The magnitude of the average emf induced in the coil is \(\mathrm{N}_{\mathrm{a}} \mathrm{BA} / \Delta \mathrm{t}\).

The magnitude of the induced emf is directly proportional to the number of turns and rate of change of magnetic field and area of the coil.