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Class 12 › Physics › Moving Charges and Magnetism
CBSE2020Class Class 12 · Physics2 Marks · Short✅ Verified
A long straight solid conductor of radius 5 cm carries a current of 2 A uniformly distributed over its circular cross-section. Find the magnetic field induction at a distance of 3 cm from the axis of the conductor.
✅ Answer & Solution
Point is inside the conductor ($r = 3$ cm < a = 5$ cm).
Using Ampere's law for inside:
$$B = \frac{\mu_0 I r}{2\pi a^2}$$
$$B = \frac{4\pi \times 10^{-7} \times 2 \times 0.03}{2\pi \times (0.05)^2}$$
$$B = \frac{4\pi \times 10^{-7} \times 0.06}{2\pi \times 0.0025} = \frac{2.4 \times 10^{-8}}{1.571 \times 10^{-2}}$$
$$B = \frac{4 \times 10^{-7} \times 0.06}{0.0025 \times 2} = \frac{\mu_0 I r}{2\pi a^2} = 4.8 \times 10^{-6} \text{ T}$$
