Question

derive an expression with the help of a siutable

diagram for the magnetic field at a point, on the

equatorial line of a bar magnet​

Answer 1

${E}_{net} = 2E\:Cos{\theta}$

$Cos{\theta} = \dfrac{l}{r'}$

${E}_{net} = \dfrac{2kq}{{r'}^{2}}\times\dfrac{l}{r'}$

r' = √l² + r²

${E}_{net} = \dfrac{k(2ql)}{{r'}^{3}}}$

As we know that (2ql) = P(dipole moment).

${E}_{net} = \dfrac{kP}{{r}^{2}+{l}^{2}}^{\frac{3}{2}}$

${E}_{net} = \dfrac{kP}{{r}^{3}}$ (If l<<r)

$\vec{E} = \dfrac{-kP}{{r}^{3}}$