Question

Derive the expression for magnetic field intensity due to a bar magnet on the axial line

Answer 1

Answer:

The magnetic field of a bar magnet on the axis is given by the expression $\frac{\mu_{0} \cdot 2 m}{4 \pi_{1} r^{3}}$

Explanation:

Let there be a unit North Pole placed at any point the magnetic forces acting as per coulomb’s law are

$\mathrm{F}_{\mathrm{n}}=\frac{\mu_{0} \cdot q_{m}}{4 \pi \cdot(r-l)^{2}}$

Similarly at the south pole the force exerted would be as in equation,

$\mathrm{F}_{\mathrm{S}}=\frac{\mu_{0} \cdot q_{m}}{4 \pi \cdot(r+l)^{2}}$

Thus the strength at point B will be difference in force experienced at north pole and south pole

$\mathrm{F}_{\mathrm{n}}-\mathrm{F}_{\mathrm{S}}=\frac{\mu_{0} \cdot q_{m} \cdot 4 r l}{4 \pi \cdot\left(r^{2}-l^{2}\right)^{2}}$

But dipole moment m = 2l.qm

So axial magnetic field $=\frac{\mu_{0} .2 m r}{4 \pi \cdot\left(r^{2}-l^{2}\right)^{2}}$

But for bar magnet l<<<r hence $\mathrm{B}=\frac{\mu_{0} \cdot 2 m}{4 \pi \cdot r^{3}}$