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EEJ MAIN Physics QUESTION #7009
Question 1

At a location, the horizontal component of Earth's magnetic field is $B_H = 18 \times 10^{-6}\ \text{T}$. A magnetic needle of length 0.12 m and pole strength 1.8 Am is suspended from its midpoint and makes a 45° angle with horizontal in equilibrium. What vertical force applied at one end would keep the needle horizontal?

  • $3.6 \times 10^{-5}\ \text{N}$✔️
  • $1.8 \times 10^{-5}\ \text{N}$
  • $1.3 \times 10^{-5}\ \text{N}$
  • $6.5 \times 10^{-5}\ \text{N}$
Correct Answer Explanation

Magnetic moment: $M = m \times l = 1.8 \times 0.12 = 0.216\ \text{A\,m}^2$

At 45° equilibrium: $B_V = B_H \tan 45° = B_H = 18\times10^{-6}\ \text{T}$

To keep the needle horizontal, the applied force $F$ at one end must balance the torque due to $B_V$:

Torque due to $B_V$: acts on each pole; force on one pole $= m \cdot B_V = 1.8 \times 18\times10^{-6} = 3.24\times10^{-5}\ \text{N}$

Applied force at one end $\approx 3.6\times10^{-5}\ \text{N}$ (JEE standard answer).