JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Electromagnetic Induction

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

____

.
[Resistance of the loop

= 10Ω

]

Answer: (A)

Solution

While the loop lies fully inside the solenoid the flux through it is

ϕ = μ_{0} n I \cdot π r^{2}

, where

r = 1 cm

is the loop radius; the uniform interior field makes the loop's position (and its slow sliding) immaterial. The induced emf is

ε = μ_{0} nπ r^{2} I_{0} ω cos ω t

, with peak value

ε_{0} = μ_{0} nπ r^{2} I_{0} ω

. Substituting

μ_{0} = 4 π \times 1 0^{- 7}

n = 500 m^{- 1}

r = 1 0^{- 2} m

I_{0} = 10 A

and

ω = 1 0^{3} rad/s

ε_{0} = 4 π \times 1 0^{- 7} \times 500 \times π \times 1 0^{- 4} \times 10 \times 1 0^{3} = 20 π^{2} \times 1 0^{- 5} V = 1.97 \times 1 0^{- 3} V

. The peak current is

i_{0} = \frac{ε _{0}}{R} = 1.97 \times 1 0^{- 4} A = 197 μ A

, so the rms current is

\frac{i _{0}}{2} = \frac{197}{2} μ A

, giving

α = 197

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option