Friday, February 8, 2008

IIT JEE Physics Formula Revision 35. Magnetic field due to a Current

Note: vectors are shown in bold letters


1. Biot Savart Law

dB = [1/4 π ε0c²]*(i) (dl*r/r³) ... (1)
where

dB = magnetic field at point P, due to current element dl
c = speed of light
i = current
r = vector joining the current element to the point P.

[1/ ε0c²] is written as µ0 and is called the permeability of vacuum.
Its value is 4 π*10-7 T-m/A


In terms of µ0 equation (1) becomes

dB = (µ0 /4 π)*(i) (dl*r/r³) ... (2)

The magnitude of the field

dB = (µ0 /4 π)*(idl sin θ/r²)

where θ is the angle between dl and r.
the direction of the field is perpendicular to the palne containing the current element and the point P according to the rules of the cross product.


4. Magnetic field due to current in a straight wire

B = (µ0i /4 πd) (cos θ1 – cos θ2) … (4)

θ1 and θ2 and the value of θ corresponding to the lower end and the upper end respectively.

If the point P is on the perpendicular bisector of the straight wire

B = (µ0i /4 πd) [2a/√(a² +4d²)] … (5)

a = length of the wire
d = distance between the wire and point P (perpendicular distance)

If the wire is very long such that θ1 = 0 and θ2 = π.
The equation is B = (µ0i /2 πd) … (6)


7. Force between two parallel currents

dF/dl = (µ0i1i2 )/2 πd

dF/dl = force per unit length of the wire W2 due to wire W1
i1,i2 = current in wire W1 and W2 respectively
d = distance between wires kept parallel to each other.


8. Magnetic field in an axial point

B = (µ0ia²)/[2(a²+d²)3/2] .. (8)

If the point is far away from the centre d is very large compared to a

B = (µ0ia²)/2d³

If the magnetic dipole moment of due to the circular conductor with area πa² and current flowing through it is ‘i’, the µ = i πa² and B is

B = (µ0i /4π)(2µ/d³) .. (9)

10. Ampere’s law


The circulation of ∫B.dl of the resultant magnetic field along a closed plane curve is equal to µ0 times the total current crossing the area bounded by the closed curve provided the electric field inside the loop remains constant.

The circulation ∫B.dl over closed curve = µ0*I … (10)

11. Magnetic field inside a solenoid

B = µ0ni

n = number of turns per unit length along the length of the solenoid.
i = current passing through the solenoid

12. toroid

B = µ0Ni/2πr
Where
N = total number of turns
i = current in the toroid
r = distance of the point at which field is being calculated from centre of the toroid.


Join
IIT JEE academy orkut community.


http://www.orkut.co.in/Community.aspx?cmm=39291603

No comments: