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

Concept: For a short bar magnet of magnetic dipole moment $M$, the magnetic field at a distant point depends upon whether the point lies on the axial line or on the equatorial (normal bisector) line. The magnetic field on the axial line is \[ B_{\text{axial}} = \frac{\mu_0}{4\pi} \frac{2M}{r^3} \] and the magnetic field on the equatorial line is \[ B_{\text{equatorial}} = \frac{\mu_0}{4\pi} \frac{M}{r^3} \] where $r$ is the distance from the centre of the magnet.

Step 1: Interpret the percentage statement The problem states that the magnetic field at point A is $1500\%$ more than the magnetic field at point B. If a quantity is $1500\%$ more than another quantity, then \[ B_A = B_B + \frac{1500}{100}B_B \] \[ B_A = B_B + 15B_B \] \[ B_A = 16B_B \] Thus, \[ B_A = 16B_B \]

Step 2: Substitute the expressions for magnetic fields Let $r_A$ be the distance of point A from the centre of the magnet and $r_B$ be the distance of point B from the centre. Using the standard formulas, \[ \frac{\mu_0}{4\pi} \frac{2M}{r_A^3} = 16 \left( \frac{\mu_0}{4\pi} \frac{M}{r_B^3} \right) \] Canceling the common factors $\dfrac{\mu_0}{4\pi}$ and $M$, \[ \frac{2}{r_A^3} = \frac{16}{r_B^3} \] Dividing both sides by $2$, \[ \frac{1}{r_A^3} = \frac{8}{r_B^3} \] Hence, \[ r_B^3 = 8r_A^3 \] Taking cube root on both sides, \[ r_B = \sqrt[3]{8}\,r_A \] \[ r_B = 2r_A \]

Step 3: Substitute the given value Given, \[ r_A = 18\text{ cm} \] Therefore, \[ r_B = 2\times 18 \] \[ r_B = 36\text{ cm} \] Hence, the distance of point B from the centre of the magnet is \[ \boxed{36\text{ cm}} \]