JEE Main 2026 Apr 4 Shift 1, Physics Q16: Torque and Work Done in a Magnetic Field

JEE Main2026Apr 4, Shift 1Physics

An insulated wire is wound so that it forms a flat coil with $N = 200$ turns. The radius of the innermost turn is $r_{1} = 3$ cm, and of the outermost turn $r_{2} = 6$ cm. If $20$ mA current flow in it then the magnetic moment will be $α \times 1 0^{- 2}$ A $\cdot$ m $^{2}$ . The value of $α$ is _______.

A
$4.4$
B
$2.64$
C
$3.25$
D
$1.2$

Solution

Treat the spiral as a continuous distribution of turns. Linear turn density: $\frac{N}{r _{2} - r _{1}}$ turns per metre of radius. An infinitesimal ring at radius $r$ has $d N = \frac{N}{r _{2} - r _{1}} d r$ turns, each contributing area $π r^{2}$ to the magnetic moment.

$M = \int_{r_{1}}^{r_{2}} I \cdot \frac{N}{r _{2} - r _{1}} \cdot π r^{2} d r = \frac{I N π}{r _{2} - r _{1}} \cdot \frac{r _{2}^{3} - r _{1}^{3}}{3} .$

Plug in $I = 20$ mA, $N = 200$ , $r_{1} = 0.03$ m, $r_{2} = 0.06$ m:

$r_{2}^{3} - r_{1}^{3} = (216 - 27) \times 1 0^{- 6} = 189 \times 1 0^{- 6}$ m $^{3}$ , and $r_{2} - r_{1} = 0.03$ m.

$M = \frac{( 20 \times 1 0 ^{- 3} ) ( 200 ) ( 3.14 ) ( 189 \times 1 0 ^{- 6} )}{3 \times 0.03} = \frac{2373.84 \times 1 0 ^{- 6}}{9 \times 1 0 ^{- 2}} \approx 2.64 \times 1 0^{- 2}$ A $\cdot$ m $^{2}$ .

Therefore $α = 2.64.$

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JEE Main 2026 Apr 4 Shift 1, Physics Q16: Torque and Work Done in a Magnetic Field

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