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QPages 230-234: Q53: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q53 Solution: A rectangular coil of 100 turns has length 5 cm and width 4 cm. …

A rectangular coil of 100 turns has length 5 cm and width 4 cm. It is placed with its plane parallel to a uniform magnetic field and a current of 2 A is sent through the coil. Find the magnitude of the magnetic field BB if the torque acting on the coil is 0.2 N m0.2 \text{ N m}.

Step-by-Step Verified Solution

**Answer:** B=0.1 TB = 0.1 \text{ T}. **Step-by-Step Explanation:** N=100N = 100, A=0.05×0.04=0.002 m2A = 0.05 \times 0.04 = 0.002 \text{ m}^2, i=2 Ai = 2 \text{ A}, τ=0.2 N m\tau = 0.2 \text{ N m}. τ=NiAB⇒B=0.2100×2×0.002=0.5 T\tau = N i A B \Rightarrow B = \frac{0.2}{100 \times 2 \times 0.002} = 0.5 \text{ T}.

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