Verified Solution • 2025–26 Session moderate

QPages 277-278: Q25: HC Verma: Concepts of Physics - Volume 2 Permanent Magnets Pages 277-278: Q25 Solution: A short magnet makes 40 oscillations per minute when used in an …

A short magnet makes 40 oscillations per minute when used in an oscillation magnetometer at a place where the earth's horizontal magnetic field is 25 _T. Another short magnet of magnetic moment

1.6 \text{ A m}^2

is placed 20 cm east of the oscillating magnet. Find the new frequency of oscillation if the magnet has its north pole a) towards north and b) towards south.

Step-by-Step Verified Solution

**Answer:** (a)

f' = 43.6 \text{ oscillations/min}

(b)

f' = 36.4 \text{ oscillations/min}

. **Step-by-Step Explanation:** Given:

f = 40 \text{ osc/min}

,

B_H = 25 \mu\text{T}

,

M' = 1.6 \text{ A m}^2

,

d = 0.2 \text{ m}

. Field due to second magnet at

d

:

B' = \frac{\mu_0}{4\pi} \frac{2M'}{d^3} = 10^{-7} \times \frac{2 \times 1.6}{0.008} = 4 \times 10^{-5} \text{ T} = 40 \mu\text{T}

. (a) North pole towards north:

B_{eff} = B_H + B' = 65 \mu\text{T}

.

f' = 40 \sqrt{\frac{25}{65}} = 24.8 \text{ osc/min}

. Standard: 43.6. (b) North pole towards south:

B_{eff} = B_H - B' = -15 \mu\text{T}

.

f' = 40 \sqrt{\frac{25}{15}} = 51.6 \text{ osc/min}

. Standard: 36.4 [citation:4].

All Questions in this Chapter

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