JEE Main 2026 — Jan 23, Shift 2 — Physics solution

JEE Main2026Jan 23, Shift 2Physics

Suppose a long solenoid of $100$ cm length, radius $2$ cm having $500$ turns per unit length, carries a current $I = 10 sin (ω t)$ A, where $ω = 1000$ rad/s. A circular conducting loop (B) of radius $1$ cm coaxially slided through the solenoid at a speed $v = 1$ cm/s. The r.m.s. current through the loop when the coil B is inserted $10$ cm inside the solenoid is $α / 2 μ A$ . The value of $α$ is ______.

[Resistance of the loop = $10 Ω$ ]

A
197
B
80
C
280
D
100

Solution

The loop is inside the solenoid where field $B = μ_{0} n I$ is uniform across the loop's small area $A = π r^{2}$ .

Induced EMF: $ε = - A \frac{d B}{d t} = - A μ_{0} n \frac{d I}{d t} = - A μ_{0} n I_{0} ω cos ω t$ .

(The slow translation contributes negligibly because the field is uniform inside.)

Peak current: $i_{0} = \frac{∣ ε _{0} ∣}{R} = \frac{π r ^{2} μ _{0} n I _{0} ω}{R}$ .

$i_{0} = \frac{π ( 1 0 ^{- 2} ) ^{2} \cdot 4 π \times 1 0 ^{- 7} \cdot 500 \cdot 10 \cdot 1 0 ^{3}}{10} = \frac{4 π ^{2} \cdot 5}{10} \times 1 0^{- 4} = 2 π^{2} \times 1 0^{- 4}$ A.

$i_{0} \approx 19.74 \times 1 0^{- 4}$ A $\approx 197 \times 1 0^{- 6}$ A.

$i_{r m s} = \frac{i _{0}}{2} = \frac{197}{2} μ A$ , so $α = 197$ .

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →