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 $2D$ 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 $d=2D$
$\therefore B_2=\frac{1}{8}\frac{{\mu }_{0}M}{4\pi D^3}$ (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
$\alpha =15/8$