Physics Circle home Student Login

Home›Chapters›Magnetic Properties of Materials PYQs›Vibration Magnetometer›Question 10

JEE Main 2022 (27 Jul, Shift 2) · Vibration Magnetometer

Vibration Magnetometer question from JEE Main 2022 (27 Jul, Shift 2)

Q10JEE Main 2022 · 27 JUL · SHIFT 2Difficult

Vibration Magnetometer

A compass needle of oscillation magnetometer oscillates 20 times per minute at a place P\mathrm{P} of dip 30∘\text{dip } 30^\circ. The number of oscillations per minute become 10 at another place Q\mathrm{Q} of 60∘60^\circ dip. The ratio of the total magnetic field at the two places (BQ:BP)(B_Q : B_P) is:

Show answer and solution

Answer: (a)

  1. Frequency n∝BHn \propto \sqrt{B_H} and BH=Bcos⁡δB_H = B\cos\delta, so n2∝Bcos⁡δn^2 \propto B\cos\delta.
  2. nP2nQ2=BPcos⁡30∘BQcos⁡60∘⇒400100=BPBQ⋅3/21/2\dfrac{n_P^2}{n_Q^2} = \dfrac{B_P\cos 30^\circ}{B_Q\cos 60^\circ} \Rightarrow \dfrac{400}{100} = \dfrac{B_P}{B_Q}\cdot\dfrac{\sqrt{3}/2}{1/2}
  3. 4=3 BPBQ⇒BQBP=344 = \sqrt{3}\,\dfrac{B_P}{B_Q} \Rightarrow \dfrac{B_Q}{B_P} = \dfrac{\sqrt{3}}{4}
  4. So the answer is (a) 3:4\sqrt{3} : 4.
Report a mistake

More questions on Vibration Magnetometer