Question

A bar magnet of magnetic moment $M$ is placed at a distance $D$ with its axis along positive $X$ -axis. Likewise, second bar manget of magnetic moment $M$ is placed at a distance $2 D$ on positive $Y$ -axis and perpendicular to it as shown in the figure. The magnitude of magnetic field at the origin
is $| B |=\alpha\left[\frac{\mu_{0}}{4 \pi} \frac{M}{D^{3}}\right]$. The value of $\alpha$ must be
(assume $D>>l$, where $l$ is the length of magnets)

• A. $2$
• B. $\frac{15}{8}$
• C. $\frac{17}{8}$
• D. $\frac{9}{8}$

Answer 1

Answer: (B)

Magnetic field at axial point $(d>>l)$
$B _{1} =\frac{2 \mu_{0} M}{4 \pi d^{3}}$ (from $S$ to $N$ )
$ =\frac{2 \mu_{0} M}{4 \pi D^{3}} (d=D)$
Magnetic field at equatorial line $(d>>l)$
$B _{2}=\frac{2 \mu_{0} M}{4 \pi d^{3}}$ (from $N$ to $s$)
Now, for equatorial point $d=2 D$
$\therefore B _{2}=\frac{1}{8} \frac{\mu_{0} M}{4 \pi D^{3}} $ (from $N$ to $s$)
So, resultant magnetic field, ( from $S$ to $N$)
$B = B _{1}- B _{2}$ ( from $S$ to $N$)
$=\frac{\mu_{0} M}{4 \pi D^{3}}\left(2-\frac{1}{8}\right)$
$B =\frac{15}{8} \frac{\mu_{0} M}{4 \pi D^{3}}$ (in magnitude)
On comparing with given formula, we get
$\alpha=15 / 8$