## Friday, February 8, 2008

### IIT JEE Physics Formula Revision 36. Permanent Magnets

1. force

A magnetic charge m placed in a magnetic field B experiences a force.

F = mB .. (1)

2. Magnetic field due to magnetic charge

B = (µ0/4 π)(m/r²)

3. Pole strength due to current

m = IA ...(3)

Where
m = pole strength
I = surface current per unit length of the magnet
A = cross sectional of the magnet

4. Magnetic moment of a bar magnet

M = 2ml … (4)

Where
M = Magnetic moment of a bar magnet
M = pole strength
2l = magnetic length of the bar magnet

5. Potential energy at an angle θ

U(θ) = -MB cos θ = -M.B …(5)

6. Magnetic field due to a bar magnet

End on position. A position on the magnetic axis of a bar magnet is called an end on position

B = (µ0/4 π)[(2Md)/(d² – l²)²] …(6)

If d is very large compared to l, then

B = (µ0/4 π)(2M/d³) … (7)

B = (µ0/4 π)[m 2l/(d² + l²)3/2 ] …(8)
= (µ0/4 π)[M/(d² + l²)3/2 ]

If d is very large compared to l,

B = (µ0/4 π)[M/(d³)

10. Magnetic scalar potential

V(r2) – V(r1) = -r1r2 B.dr … (10)

11. The component of the magnetic field in any direction is given by

Bl = -dV/dl ... (11)

12.For a pole of pole strength m, the field at a distance r is

B = (µ0/4 π)[m/r²)

So the potential at a distance r is

V( r )= - r 0/4 π)[m/r²)dr

= (µ0/4 π)[m/r) ... (12)

13. Magnetic scalar potential due to a magnetic dipole

Magnetic scalar potential at a point P which is at a distance r from the mid point of the magnetic dipole, and the angle between the dipole axis and the line joining the mid point of the dipole to the point P is θ

V = (µ0/4 π)[Mcos θ /r²)
Where
M = 2ml =magnetic moment of the dipole

14. Magnetic field due to dipole

Magnetic field at P =

0/4 π)[M /r²)√(1 +3 cos² θ)] .. (14)

17. current in galvanometer

i = K tan θ .. (17)

where K = 2rBH0 for the given galvanometer at a given place.

18. Current in moving coil galvanometer

i = (k/nAB) θ … (18)

the constant (k/nAB) is called the galvanometer constant and may be found by passing a known current, measuring the deflection θ and putting these values in equation (18).

19 Shunt
Shunt is a small resistance.
Current in galvanometer

ig = [Rs/(Rs+g)]*i

i = main current
ig = current that goes through galvanometer

20. tangent law of perpendicular fields

B = BHtan θ .. (20)

21. Deflection magnetometer

Tan A position

M/ BH = (4 π/µ0)[(d² – l²)²/2d] tan θ

22. Deflection magnetometer Tan B position

M/ BH = (4 π/µ0)[(d² – l²)3/2] tan θ.. (22)

23. Oscillation Magnetometer

Time period T = 2 π/ ω = 2 π√ (I/M BH) .. (23)
From equation (23)

M BH) = 4 π²I/T² … (24)

25. Gauss’s law for magnetism: the flux of the magnetic field through any closed surface is zero

integral over the closed surface ∫B.ds = 0 .. (25)

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