Question

Obtain an expression for magnetic field due to bar magnet at a point along its axis. A short magnetic magnetic dipole has magnetic moment 0.7Am^2. Calculate its axial magnetic field at a distance of 10cm from center of magnetic dipole ( Given : 4n× 10^-7. S.I.unit)​

Answer 1

Answer:

Consider a point P on equatorial (or broad side on position) of short bar magnet of length 2I, having north pole (N) and south pole (S) of strength +q

m and −qm

respectively. The distance of point P from the mid-point (O) of magnet is r. Let

B1 and B2

be the magnetic field intensities due to north and south poles respectively. NP=SP=r2+l^2.

vecB1=4πμ0r2+l^2qm

along N to P

vecB2=4πμ0r2+l^2

qm

along P to S

Clearly, magnitude of and are equal

i.e., ∣B^1

∣=∣B2

∣ or B1=B2

To find the resultant of

B1 and B2

we resolve them along and perpendicular to magnetic axis SN components

B1

of along and perpendicular to magnetic axis are B1

cosθ and B2

sinθ respectively.

Components of

B2

along and perpendicular to magnetic axis are 2cosθ and B2sinθ respectively. Clearly components of and perpendicular to axis SN. B1sinθ and B2sinθ ae equal in magnitude and opposite in direction and hence, cancel each other, while the components of

B1 and B2

along the axis are in the same direction and hence, add up to give to resultant magnetic filed parallel to the direction

NS.

∴ Resultant magnetic field intensity at P.

B=B1

cosθ+B2

cosθ

But B1=B2=4πμ0r2+l^2qm

ancosθ=PNON=r2+l21=

(r2+l2)1/2

1∴ B=2B1cosθ=2×4πμ0

(r2+l2)qm×

(r2+l2)1/2

l=4πμ0

(r2+l2)

3/2

2qm^l

But qm

2I=m, magnetic moment of magnet∴ B=4πμ0(r2+l2)3/2m

Explanation:

hope this helps you out