JEE Main 2025 Jan 29 Shift 1, Physics Q9: Magnetic Field due to Electric Current

JEE Main2025Jan 29, Shift 1Physics

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 anywhere due to the wire, will be

A
$[a /4, 3 a /2]$
B
$[a /2, 2 a]$
C
$[a /2, 3 a]$
D
$[a /4, 2 a]$

Solution

Using Ampere's law, the field inside the wire grows linearly with $r$ and outside it falls as $1/ r$ :

$B_{in} (r) = \frac{μ _{0} I r}{2 π a ^{2}} (r \leq a), B_{out} (r) = \frac{μ _{0} I}{2 π r} (r \geq a) .$

Both reach the same value at $r = a$ , which is the maximum: $B_{m a x} = \frac{μ _{0} I}{2 π a} .$

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

Inside: $\frac{μ _{0} I r}{2 π a ^{2}} = \frac{μ _{0} I}{4 π a} \Rightarrow r = a /2.$

Outside: $\frac{μ _{0} I}{2 π r} = \frac{μ _{0} I}{4 π a} \Rightarrow r = 2 a .$

$∴ [a /2, 2 a] .$

Concept behind this question

Magnetic Field due to Electric Current Notes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 29 Shift 1, Physics Q9: Magnetic Field due to Electric Current

Concept behind this question

More previous year questions on Magnetic Field due to Electric Current

Practice more Physics