Q126 / 180 · COMEDK 2026 (9 May Shift 2) · Physics
On both sides of a magnetic needle, two short magnets A and B are placed on the same horizontal line which is perpendicular to the magnetic meridian. The south poles of A and B are facing each other, which are 10 cm and 20 cm respectively from the magnetic needle. If the needle remains undeflected, the ratio of the magnetic moment of A to that B is:
- A$2:1$
- B$1:2$
- C$8:1$
- Correct answerD$1:8$
Solution
The magnetic field due to a short magnet on its axial line at a distance $d$ is given by $B = \dfrac{\mu_0}{4\pi} \dfrac{2M}{d^3}$.
Since the magnetic needle remains undeflected, the magnetic fields produced by magnets A and B at the position of the needle must be equal and opposite.
$B_A = B_B$
$\dfrac{\mu_0}{4\pi} \dfrac{2M_A}{d_A^3} = \dfrac{\mu_0}{4\pi} \dfrac{2M_B}{d_B^3}$
$\dfrac{M_A}{M_B} = \left(\dfrac{d_A}{d_B}\right)^3$
Given $d_A = 10\text{ cm}$ and $d_B = 20\text{ cm}$.
$\dfrac{M_A}{M_B} = \left(\dfrac{10}{20}\right)^3 = \dfrac{1}{8}$
The ratio of the magnetic moment of A to that of B is $1:8$.
Answer: $1:8$
Since the magnetic needle remains undeflected, the magnetic fields produced by magnets A and B at the position of the needle must be equal and opposite.
$B_A = B_B$
$\dfrac{\mu_0}{4\pi} \dfrac{2M_A}{d_A^3} = \dfrac{\mu_0}{4\pi} \dfrac{2M_B}{d_B^3}$
$\dfrac{M_A}{M_B} = \left(\dfrac{d_A}{d_B}\right)^3$
Given $d_A = 10\text{ cm}$ and $d_B = 20\text{ cm}$.
$\dfrac{M_A}{M_B} = \left(\dfrac{10}{20}\right)^3 = \dfrac{1}{8}$
The ratio of the magnetic moment of A to that of B is $1:8$.
Answer: $1:8$
Hence, the correct answer is option: 'D'.