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# Physics Project Report on Torque on a Current Carrying Coil Placed in a Uniform Magnetic Field (M.F.)

Contents

Certificate

Acknowledgement

i) Principle

ii) Construction

iii) Force Responsible for producing Torque

iv) Bibliography

CERTIFICATE

This is to certify that the investigatory Project Report entitled "Torque on a Current Carrying Coil Placed in a Uniform Magnetic Field (M.F.)"submitted by student of Class XII is original and has been completed by him under my supervision.

ACKNOWLEDGEMENT

As a student of Class XII. I did this project on a a part of my studies entitled to "Torque on a Current Carrying Coil Placed in a Uniform Magnetic Field (M.F.)". I owe a deep sense of gratitude to my Physics Teacher whose valuable advice, guidance and helped me in doing this project from conception to completion.

At the same time, I can not forget to express my thankfulness to our school Principal for extending his generous, patronage and constant encouragement.

Finally I thankful my parents for helping me economically and my friends for giving me a helping hand at every step of the project.

Signature of the Student

Introduction

Principle : - It is based upon the Principle that a net torque will act on a current carrying coil and this torque will tends to rotate the coil.

Construction :- It consist of a rectangular Coil (ABCD). Such that AB = CD = l and AD = BC = b. Now let current (I) will flows through it and it is be putted in uniform Magnetic field also the normal of the coil makes an angle (q) with magnetic field (M.F.)

Note : (i) Here (q) be the angle between the normal of the coil and the magnetic field. But (µ) is the angle between the coil and the magnetic field.

(ii) Taking (q) i.e. in force and it makes (90+q) in and (90-q) in .

Let F1, F2 & F3, F4 be the forces acting on the sides DA, AB, BC and DA respectively.

The force F4 acting on arm DA is

F1 = I (DA) B Sin (90+q) = I b B Cos q -----------> (1)

The force F1 acting on arm DA in the plane of the coil and in up ward direction.

Now let F3 bc the force acting on the arm (BC) so (90-q)

The angle between the normal on the coil and the magnetic field. So , F3 = I (BC) B Sin (90-q) = I b B Cos -----------> (2)

Its direction is downward to the plane of coil.

Since F1 and F3 are the forces of equal magnitude but we not that the line of force acting on the coil is same. So they cancel the effect of each other. So they these forces F1 and F3 are not so responsible for rotating the coil.

## Forces Responsible for Producing Torque

Now : - Let F2 be the force acting on the arm AB, and (µ) be the angle between the plane of the coil and the M.F.

So F2 = I (AB) B Sin µ = I l B Sin µ

But µ = 900 so Sin 900 = 1

F2 = I l B

Let F4 be the forces acting on arm (CD)

F4 = I (CD) B Sin µ = I l B Sin 900 = 1

So we see that F2 and F4 are the equal forces and the point to be noted that these two equal forces are on opposite line of action. So it results in producing the torque.

Magnitude of (t) is given by

t = Force * ^ distance

But force = I l B and ^ distance = KD

t = I l B * KD

Note :- i) ĐKAD = q = b Sin q

Now : t = IAB Sin q

Special Cases : (i) If the magnetic field makes an angle (µ) with the plane of coil then (q+ µ) = 900 = q = 90-q)

t = IAB Sin (90-q) =

t = IAB Cos µ

ii) If the coil consists of (n) number of turns then

t = nIAB Sin q

or t = nIAB Cos µ

Now :- i) We take Sin q when angle is beteen the normal of the coil and M.F.

ii) We take Cos µ, when angle is between the plane of coil and M.F

Bibliography

1. Modern’s abc of Physics.

3. Dinesh a to z Physics.

4. Internet Sources : www.google.com, www.wikipedia.in