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Class 12 › Physics › Moving Charges and Magnetism
CBSE2021Class Class 12 · Physics1 Marks · Mcq✅ Verified
A long straight wire of circular cross-section of radius $a$ carries a steady current $I$ uniformly distributed across its cross-section. The ratio of magnitudes of magnetic field at a point $a/2$ above the surface to that at a point $a/2$ below its surface is
A. 04:01
B. 01:01
C. 04:03
D. 03:04
✅ Answer & Solution
✅ Correct Answer: C
Point above surface: distance from axis $r_1 = a + a/2 = 3a/2$ (outside wire).
$$B_1 = \frac{\mu_0 I}{2\pi r_1} = \frac{\mu_0 I}{2\pi \cdot 3a/2} = \frac{\mu_0 I}{3\pi a}$$
Point below surface: distance from axis $r_2 = a - a/2 = a/2$ (inside wire).
$$B_2 = \frac{\mu_0 I r_2}{2\pi a^2} = \frac{\mu_0 I \cdot a/2}{2\pi a^2} = \frac{\mu_0 I}{4\pi a}$$
$$\frac{B_1}{B_2} = \frac{\mu_0 I / 3\pi a}{\mu_0 I / 4\pi a} = \frac{4}{3}$$
