Question

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 $10\text{ cm}$ and $20\text{ cm}$ respectively from the magnetic needle. If the needle remains undeflected, the ratio of the magnetic moment of $A$ to that of $B$ is:

• A. $1:8$
• B. $2:1$
• C. $8:1$
• D. $1:2$

Hint:

For balancing fields on a common axis, the magnetic moment is directly proportional to the cube of the distance ($M \propto d^3$). Since the distance of magnet $B$ is double that of $A$, its magnetic moment must be $2^3 = 8$ times larger to maintain equilibrium!

Answer 1

The Correct Option is A

Understanding the Concept: This setup describes a deflection magnetometer arranged in the End-on position (or Tan-A position). The magnetic field ($B$) on the axial line of a short bar magnet of magnetic moment $M$ at a distance $d$ from its center is given by: \[ B = \frac{\mu_0}{4\pi} \frac{2M}{d^3} \implies B \propto \frac{M}{d^3} \] For the magnetic needle to remain completely undeflected, the individual axial fields produced by magnets $A$ and $B$ at the center must balance each other out exactly ($B_A = B_B$).
Step 1: Set up the magnetic field balance equation.
Equating the two field expressions: \[ \frac{\mu_0}{4\pi} \frac{2M_A}{d_A^3} = \frac{\mu_0}{4\pi} \frac{2M_B}{d_B^3} \implies \frac{M_A}{d_A^3} = \frac{M_B}{d_B^3} \] Rearranging terms to isolate the ratio of magnetic moments: \[ \frac{M_A}{M_B} = \left(\frac{d_A}{d_B}\right)^3 \]
Step 2: Substitute the given distance values.
Given distances $d_A = 10\text{ cm}$ and $d_B = 20\text{ cm}$: \[ \frac{M_A}{M_B} = \left(\frac{10}{20}\right)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \implies 1:8 \]