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QPages 277-278: Q7: HC Verma: Concepts of Physics - Volume 2 Permanent Magnets Pages 277-278: Q7 Solution: A bar magnet has a length of 8 cm. The magnetic field at a…

A bar magnet has a length of 8 cm. The magnetic field at a point at a distance 3 cm from the centre in the broadside-on position is found to be 4×10−6 T4 \times 10^{-6} \text{ T}. Find the pole strength of the magnet.

Step-by-Step Verified Solution

**Answer:** m=6.25×10−2 A mm = 6.25 \times 10^{-2} \text{ A m}. **Step-by-Step Explanation:** Given: 2l=8 cm2l = 8 \text{ cm}, d=3 cmd = 3 \text{ cm}, B=4×10−6 TB = 4 \times 10^{-6} \text{ T}. Broadside-on: B=μ04πm(2l)(d2+l2)3/2B = \frac{\mu_0}{4\pi} \frac{m(2l)}{(d^2 + l^2)^{3/2}}. 4×10−6=10−7×m×0.08(0.0009+0.0016)3/24 \times 10^{-6} = 10^{-7} \times \frac{m \times 0.08}{(0.0009 + 0.0016)^{3/2}}. Solving gives m=6.25×10−2 A mm = 6.25 \times 10^{-2} \text{ A m} [citation:17].

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