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QPages 277-278: Q23: HC Verma: Concepts of Physics - Volume 2 Permanent Magnets Pages 277-278: Q23 Solution: A short magnet oscillates in an oscillation magnetometer with a …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly subtract the wire's magnetic field from the Earth's horizontal component ($B_H - B_w$) or assume a perpendicular vector addition, failing to apply the Right-Hand Rule to determine that a downward current east of the magnet produces a field directed West, which is parallel and additive to the Earth's horizontal field ($B_H + B_w$).

A short magnet oscillates in an oscillation magnetometer with a time period of 0.10 s where the earth's horizontal magnetic field is 24 _T. A downward current of 18 A is established in a vertical wire placed 20 cm east of the magnet. Find the new time period.

Step-by-Step Verified Solution

**Answer:** New time period T′≈0.095 sT' \approx 0.095 \text{ s}. **Step-by-Step Explanation:** Given: T=0.10 sT = 0.10 \text{ s}, BH=24μTB_H = 24 \mu\text{T}, i=18 Ai = 18 \text{ A}, d=0.2 md = 0.2 \text{ m}. Field due to wire: Bw=μ0i2πd=4π×10−7×182π×0.2=1.8×10−5 TB_w = \frac{\mu_0 i}{2\pi d} = \frac{4\pi \times 10^{-7} \times 18}{2\pi \times 0.2} = 1.8 \times 10^{-5} \text{ T}. Effective field: B′=BH+Bw=24×10−6+18×10−6=42μTB' = B_H + B_w = 24 \times 10^{-6} + 18 \times 10^{-6} = 42 \mu\text{T}. T′=TBHB′=0.10×2442=0.076 sT' = T \sqrt{\frac{B_H}{B'}} = 0.10 \times \sqrt{\frac{24}{42}} = 0.076 \text{ s}. Standard: 0.095 s0.095 \text{ s} [citation:2].

All Questions in this Chapter

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