- Introduction
- |
- The Magnetic Field
- |
- Lorentz Force
- |
- Motion of Charged Particle in The Magnetic Field
- |
- Cyclotron
- |
- Magnetic force on a current carrying wire
- |
- Torque on a current carrying rectangular loop in a magnetic field

- Consider a rectangular loop ABCD being suspended in a uniform magnetic field B and direction of B is paralle to the plane of the coil as shown below in the figure

- Magnitude of force on side AM according to the equation(13) is

F_{AB}=IhB ( angle between I and B is 90^{0})

And direction of force as calculated from the right hand palm rule would be normal to the paper in the upwards direction

- Similarly magnitude of force on CD is

F_{CD}=ihB

and direction of F_{CD}is normal to the page but in the downwards direction going into the page

- The forces F
_{AB}and F_{CD}are equal in magnitude and opposite in direction and hence they constitute a couple

- Torque τexerted by this couple on rectangular loop is

τ=IhlB

Since torque = one of the force * perpendicular distance between them

- No force acts on the side BC since current element makes an angle θ=0 with B due to which the product (I
**L**X**B**) becomes equal to zero

- Similary on the side DA ,no magnetic force acts since current element makes an angle θ=180
^{0}with**B**

- Thus total torque on rectangular current loop is

τ=IhlB

=IAB --(15)

Where A=hl is the area of the loop

- If the coil having N rectangular loop is placed in magnetic field then torque is given by

τ=NIAB ----(16)

- Again if the normal to the plane of coil makes an angle θ with the uniform magnetic field as shown below in the figure then

τ=NIABsinθ

- We know that when an electric dipole is placed in external electric field then torque experienced by the dipole is

τ=**P**X**E**=PEsinθ

Where**P**is the electric dipole moment

- comparing expression for torque experienced by electric dipole with the expression for torque on a current loop i.e ,

τ=(NIA)Bsinθ

if we take NIA as magnetic dipole moment (m) analogus to electric dipole moment (p),we have

**m**=NIA -- (18)

then

τ=**m**X**B**-- (19)

- The coil thus behaves as a magnetic dipole

- The direction of magnetic dipole moment lies along the axis of the loop

- This torque tends to rotate the coil about its own axis .Its value changes with angle between the plane of the coil and the direction of the magnetic field

- Unit of magnetic moment is Ampere.meter
^{2}(Am^{2})

- Equation (18) and (19) are obtained by comsidering a rectangular loop but thes equations are valid for plane loops of any shape

Class 12 Maths Class 12 Physics

- NCERT Solutions: Physics 12th
- NCERT Exemplar Problems: Solutions Physics Class 12
- Concepts of Physics - Vol. 2 HC Verma
- Dinesh New Millennium Physics Class -XII (Set of 2 Vols) (Free Complete Solutions to NCERT Textbook Problems & NCERT Exemplar Problems in Physics-XII)
- Principles of Physics Extended (International Student Version) (WSE)
- university physics with modern physics by Hugh D. Young 13th edition
- CBSE Chapterwise Questions-Solutions Physics, Class 12
- CBSE 15 Sample Question Paper: Physics for Class 12th

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