Question

The magnetic field at a point A on the axis of a short bar magnet is $1500\%$ more than the magnetic field at a point B on the normal bisector of the magnet. If the distance of point A from the centre of the magnet is $18\text{ cm}$, then the distance of point B from the centre of the magnet is:

• A. $72\text{ cm}$
• B. $36\text{ cm}$
• C. $48\text{ cm}$
• D. $54\text{ cm}$

Hint:

For a short bar magnet, \[ B_{\text{axial}} = \frac{\mu_0}{4\pi}\frac{2M}{r^3} \] and \[ B_{\text{equatorial}} = \frac{\mu_0}{4\pi}\frac{M}{r^3} \] Always remember that the axial field is twice the equatorial field at the same distance from the centre. Also, "$1500\%$ more" means multiplying the original value by $16$, not by $15$.

Answer 1

The Correct Option is B

Step 1: Recall the two short-magnet fields.
On the axis, $B_{axial} = \dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}$, and on the normal bisector (equatorial line), $B_{equatorial} = \dfrac{\mu_0}{4\pi}\dfrac{M}{r^3}$.
Step 2: Translate the percentage.
Point A's field is $1500\%$ more than B's, meaning $B_A = B_B + 15B_B = 16B_B$.
Step 3: Insert the field expressions.
$\dfrac{\mu_0}{4\pi}\dfrac{2M}{r_A^3} = 16\left(\dfrac{\mu_0}{4\pi}\dfrac{M}{r_B^3}\right)$.
Step 4: Cancel common factors.
Dropping $\dfrac{\mu_0}{4\pi}$ and $M$, $\dfrac{2}{r_A^3} = \dfrac{16}{r_B^3}$, so $\dfrac{1}{r_A^3} = \dfrac{8}{r_B^3}$, giving $r_B^3 = 8 r_A^3$.
Step 5: Take the cube root.
$r_B = \sqrt[3]{8}\,r_A = 2 r_A$.
Step 6: Substitute and conclude.
With $r_A = 18\,cm$, $r_B = 2\times 18 = 36\,cm$, which is option (2).
\[ \boxed{r_B = 36\,cm} \]