# You hold a coil consisting of 20 turns of wire around the periphery of a rectangular frame 10 \...

## Question:

You hold a coil consisting of {eq}20 {/eq} turns of wire around the periphery of a rectangular frame {eq}10 \ cm {/eq} by {eq}5.0 \ cm {/eq}. You move it from a position where the average magnetic field is {eq}1.0 \ mT {/eq} to another position where the average is {eq}11 \ mT {/eq}. If the plane of the coil is perpendicular to average magnetic field and this motion takes place in {eq}1.0 \ s {/eq}, the average EMF induced in the coil is:

a. {eq}0.10 \ kV {/eq}

b. {eq}1.0 \ kV {/eq}

c. {eq}1.0 \ mV {/eq}

d. {eq}0.10 \ mV {/eq}

e. {eq}10 \ mV {/eq}

Do you have to push, pull or exert on force in the process of moving the coil?

## Rectangular Coil Moving Through a Non Uniform Magnetic Field

A non uniform magnetic field is the one in which strength of magnetic field varies from place to place. When a rectangular coil of area *A* moves through a such a non uniform magnetic field, the magnetic flux through the coil changes. The magnetic flux through the coil can be expressed as {eq}\phi = B A \cos \theta {/eq}. Here {eq}B, \ \ \theta {/eq} are the non uniform magnetic field and angle between the magnetic field and the normal from the plane of the coil. According to Faraday's law rate of change of magnetic flux with respect to time gives the magnitude of EMF induced in the coil.

- Induced EMF in the coil when it moves through a nonuniform magnetic field {eq}E = - \dfrac { \Delta \phi } { \Delta t } = - \dfrac { ( B_2 - B_1 ) A \cos \theta } { \Delta t } {/eq}. In this equation {eq}B_1, \ \ B_2, \ \ \Delta t {/eq} are the magnetic field at the initial and final positions of the coil and time taken for the coil to move from initial position to final position.

- The negative sign in the equation comes from Lenz's law and it indicates that the EMF and current induced opposes the cause producing it.

## Answer and Explanation:

**Given data**

- Number of turns in the coil {eq}N = 20 {/eq}

- Area of the coil {eq}A = 0.10 \times 0.05 = 5.0 \times 10^{-3 } \ m^2 {/eq}

- Average magnetic field at the initial position {eq}B_1 = 1.0 \times 10^{-3} \ T {/eq}

- Average magnetic field at the final position {eq}B_2 = 11.0 \times 10^{-3} \ T {/eq}

- The plane of the coil is perpendicular to the magnetic field , therefore angle between the magnetic field and normal from the plane of the coil {eq}\theta = 0 ^o {/eq}

- Time taken for the coil to move from initial position to final position {eq}\Delta t = 1.0 \ s {/eq}

**Part a )**

Magnitude of the average EMF induced in the coil {eq}V = \dfrac { \Delta \phi } { \Delta t } \\ V = \dfrac {N \times ( B_2 - B_1 ) \times A } { \Delta t } \\ V = \dfrac { 20 \times ( 11.0 \times 10^{-3 } - 1.0 \times 10^{-3 } ) \times 5.0 \times 10^{-3 } } { 1 .0 } \\ V = 1.0 \times 10{-3} \ V \\ V = 1.0 \ m V {/eq}

Therefore the correct option is **option c)**

**Part b)**

When the rectangular coil is at the initial position no magnetic force acts on it because the coil is at rest and no current flows through it.

But when the coil starts moving with some constant velocity, a change in magnetic flux and induced EMF and current occurs in the coil.

As a result, each sides of the coil experience magnetic force. The forces acting on opposite sides of the coil are directed opposite to each other.

As a result force exerted by the horizontal sides of the coil cancel each other.

But in the case of two vertical sides, the force are opposite in direction but does not cancel each other. Because the front end of the coil is at a higher magnetic field than the rear end of the coil.

So we need to apply some constant force to move the coil that is equal to the difference in force acting on the two vertical sides of the coil to move with constant velocity through a non uniform magnetic field normal to the plane of the coil.

#### Learn more about this topic:

from High School Physics: Help and Review

Chapter 13 / Lesson 10