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QPages 286-287: Q4: HC Verma: Concepts of Physics - Volume 2 Magnetic Properties of Matter Pages 286-287: Q4 Solution: A bar magnet of length 1 cm and cross-sectional area produces a …

A bar magnet of length 1 cm and cross-sectional area 1.0 cm21.0 \text{ cm}^2 produces a magnetic field of 1.5×10−4 T1.5 \times 10^{-4} \text{ T} at a point in end-on position at a distance 15 cm away from the centre. a) Find the magnetic moment M of the magnet. b) Find the magnetisation I of the magnet. c) Find the magnetic field B at the centre of the magnet.

Step-by-Step Verified Solution

**Answer:** (a) M=2.5 A m2M = 2.5 \text{ A m}^2 (b) I=2.5×106 A m−1I = 2.5 \times 10^6 \text{ A m}^{-1} (c) B≈3.14 TB \approx 3.14 \text{ T}. **Step-BY-Step Explanation:** Given: l=1 cm=0.01 ml = 1 \text{ cm} = 0.01 \text{ m}, A=1 cm2=10−4 m2A = 1 \text{ cm}^2 = 10^{-4} \text{ m}^2, d=15 cm=0.15 md = 15 \text{ cm} = 0.15 \text{ m}, B=1.5×10−4 TB = 1.5 \times 10^{-4} \text{ T}. (a) For end-on position: B=μ04π2Md(d2−l2)2B = \frac{\mu_0}{4\pi} \frac{2Md}{(d^2 - l^2)^2}. Approximating d≫ld \gg l: B≈μ04π2Md3B \approx \frac{\mu_0}{4\pi} \frac{2M}{d^3}.
1.5×10−4=10−7×2M(0.15)3⇒M=2.5 A m21.5 \times 10^{-4} = 10^{-7} \times \frac{2M}{(0.15)^3} \Rightarrow M = 2.5 \text{ A m}^2
(b) Volume V=A×l=10−4×0.01=10−6 m3V = A \times l = 10^{-4} \times 0.01 = 10^{-6} \text{ m}^3. I=M/V=2.5/10−6=2.5×106 A/mI = M/V = 2.5 / 10^{-6} = 2.5 \times 10^6 \text{ A/m}. (c) Inside the magnet, B=μ0(H+I)≈μ0I=4π×10−7×2.5×106≈3.14 TB = \mu_0 (H + I) \approx \mu_0 I = 4\pi \times 10^{-7} \times 2.5 \times 10^6 \approx 3.14 \text{ T}.

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