For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its ce
For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its center is given by `B=(mu_0IR^2N)/(2(x^2+R^2)^(3//2))`
(a) Show that this reduces to the familiar result for field at the centre of the coil.
(b) Consider two parallel coaxial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R and is given by `B=0*72(mu_0NI)/(R)` approximately.
[Such as arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]
1 Answers
Radius of circular coil `= R`
Number of turns on the coil `=N`
Current in the coil `=I`
Magnetic field at a point its axis at distance x is given by the relation,
`B = (mu_(0)IR^(2)N)/(2(x^(2)+R^(2))^(3/2))`
Where,
`mu_(0) =` Permeabilty of free space
(a) If the magnetic field at the centre of the coil is considered, then `x = 0`.
`:. B =(mu_(0)IR^(2)N)/(2R^(3)) =(mu_(0)IN)/(2R)`
This is the familar result for magnetic field at the centre of the oil.
(b) Radii of two parallel co-axial circular coils `=R`
Number pf turms on each coil = N
Current in both coils `=I`
Distance between both the coils `=R`
Let us consider point Q at distance d from the centre.
Then, one coil is at a distance of `(R)/(2)+d` from point Q.
Magnetic field at point Q is give as:
`B_(1) = (mu_(0)NIR^(2))/(2[((R)/(2)+d)^(2)+R^(2)]^(3//2))`
Also, the other coil is at a distance of `(R)/(2)-d` from point Q.
`:.` Magnetic field due to this coil is given as:
`B_(2) = (mu_(0)NIR^(2))/(2[((R)/(2)+d)^(2)+R^(2)]^(3//2))`
Total magnetic field,
`B = B_(1)+B_(2)`
`=(mu_(0)IR^(2))/(2)[{((R)/(2)-d)^(2)+R^(2)}^(3/2)+{((R)/(2)+d)^(2)+R^(2)}^(-(3)/(2))]`
`=(mu_(0)IR^(2))/(2)[((5R^(2))/(4)+d^(2)-Rd)^(-(3)/(2))+((5R^(2))/(2)+d^(2)+Rd)^(-(3)/(2))]`
`=(mu_(0)IR^(2))/(2)xx((5R^(2))/(4))^(-(3)/(2))[(1+(4)/(5)(d^(2))/(R^(2))-(4)/(5)(d)/(R))^(-(3)/(2))+(1+(4)/(5)(d^(2))/(R^(2))+(4)/(5)(d)/(R))^(-(3)/(2))]`
For `d lt lt R`, neglecting the factor `(d^(2))/(R^(2))`, we get:
`~~(mu_(0)IR^(2))/(2) xx ((5R^(2))/(4))^(-(3)/(2))xx[(1-(4d)/(5R))^(-(3)/(2))+(1+(4d)/(5R))^(-(3)/(2))]`
`~~(mu_(0)IR^(2)N)/(2R^(3)) xx ((4)/(5))^(3/(2))[1-(6d)/(5R)+1+(6d)/(5R)]`
`B = ((4)/(5))^(3//2) (mu_(0)IN)/(R) = 0.72 ((mu_(0)IN)/(R))`
Hence, it is proved that the field on the axis around the mid-point between the coils is uniform.