MEC O11

Solution Question 11

Solution for Question 11

Analysis: We need to express the magnetic induction vector $\vec{B}$ at a general point $P(r, \theta)$ in terms of radial ($\hat{e}_r$) and transverse ($\hat{e}_{\theta}$) unit vectors. This is a standard derivation for the field of a dipole.

m r θ P B_r Radial (e_r) B_θ Transverse (e_θ)

The magnetic field of a short dipole can be resolved into two components:

  • Radial Component ($B_r$): Directed along the position vector $\vec{r}$. $$ B_r = \frac{\mu_0}{4\pi} \frac{2m \cos\theta}{r^3} $$
  • Transverse Component ($B_{\theta}$): Directed perpendicular to $\vec{r}$ (in the direction of increasing $\theta$). $$ B_{\theta} = \frac{\mu_0}{4\pi} \frac{m \sin\theta}{r^3} $$

Combining these into vector form:

$$ \vec{B}_P = B_r \hat{e}_r + B_{\theta} \hat{e}_{\theta} $$ $$ \vec{B}_P = \frac{\mu_0 m}{4\pi r^3} (2\cos\theta) \hat{e}_r + \frac{\mu_0 m}{4\pi r^3} (\sin\theta) \hat{e}_{\theta} $$ $$ \vec{B}_P = \frac{\mu_0 m}{4\pi r^3} [ (2\cos\theta)\hat{e}_r + (\sin\theta)\hat{e}_{\theta} ] $$

Correct Answer: (a)

$\frac{\mu_0 m}{4\pi r^3} \{(2\cos\theta)\hat{e}_r + (\sin\theta)\hat{e}_{\theta} \}$