JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Magnetic Effects of Current

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Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire's cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be

(A) $[a /4, 3 a /2]$
(B) $[a /2, 2 a]$ ✓ Correct
(C) $[a /2, 3 a]$
(D) $[a /4, 2 a]$

Answer: (B)

Solution

For a long straight wire of radius

a

carrying current

I

uniformly over its cross-section, the field is maximum at the surface:

B_{m a x} = \frac{μ _{0} I}{2 π a}

We need the radius at which

B = B_{m a x} /2 = \frac{μ _{0} I}{4 π a}

.

Inside the wire (

r < a

), by Ampère's law the enclosed current is proportional to

r^{2} / a^{2}

, giving

B_{in} = \frac{μ _{0} I r}{2 π a ^{2}}

Setting this to

\frac{μ _{0} I}{4 π a}

\frac{μ _{0} I r}{2 π a ^{2}} = \frac{μ _{0} I}{4 π a} ⟹ r = \frac{a}{2}

Outside the wire (

r > a

B_{out} = \frac{μ _{0} I}{2 π r}

Setting this to

\frac{μ _{0} I}{4 π a}

\frac{μ _{0} I}{2 π r} = \frac{μ _{0} I}{4 π a} ⟹ r = 2 a

The required distances are

[\frac{a}{2}, 2 a]

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Magnetic Effects of Current

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