JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Magnetic Effects of Current

A coaxial cable consists of an inner wire of radius '

a

' surrounded by an outer shell of inner and outer radii '

b

' and '

c

' respectively. The inner wire carries an electric current

i_{0}

, which is distributed uniformly across cross-sectional area. The outer shell carries an equal current in opposite direction and distributed uniformly. What will be the ratio of the magnetic field at a distance

x

from the axis when (i)

x < a

and (ii)

a < x < b

(A) $\frac{x ^{2}}{a ^{2}}$ ✓ Correct
(B) $\frac{b ^{2} - a ^{2}}{x ^{2}}$
(C) $\frac{a ^{2}}{x ^{2}}$
(D) $\frac{x ^{2}}{b ^{2} - a ^{2}}$

Answer: (A)

Solution

Using Ampère's law: For

x < a

, the enclosed current is

i_{e n c} = i_{0} \frac{π x ^{2}}{π a ^{2}}

, so

B_{1} = \frac{μ _{0} i _{0} x}{2 π a ^{2}}

. For

a < x < b

, the entire inner wire current is enclosed:

B_{2} = \frac{μ _{0} i _{0}}{2 π x}

. The ratio is

\frac{B _{1}}{B _{2}} = \frac{μ _{0} i _{0} x / ( 2 π a ^{2} )}{μ _{0} i _{0} / ( 2 π x )} = \frac{x ^{2}}{a ^{2}}

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Magnetic Effects of Current

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