A closely wound coil of 100 turns and an area of cross-section 1 cm3 has a coefficient of self-induction 1 mH. The magnetic induction in the centre of the core of the coil when a current of 2A flows in it, will be:
Correct answer: A. 0.022 Wbm-2
- A. 0.022 Wbm-2
- B. 0.4 Wbm-2
- C. 0.8 Wbm-2
- D. 1 Wbm-2
Explanation
To find the magnetic induction (B) in the center of the coil, we can use the formula: B = (μ₀ * N * I) / L, where:μ₀ (permeability of free space) is approximately 4π x 10-7 H/m,N is the number of turns of the coil (100),I is the current (2 A),L is the inductance (1 mH = 1 x 10-3 H).Plugging in the values, we calculate:B = (4π x 10-7 H/m * 100 * 2 A) / (1 x 10-3 H) = 0.022 Wbm-2.This confirms that Option A is the correct answer. The other options do not match the calculated magnetic induction based on the values given in the question, hence they are incorrect.
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