JEE Main 2026 — Physics Electromagnetism

2026mcqmedium

A square loop of side $2$ cm is placed in a time varying magnetic field with magnitude as $B = 0.4\sin(300t)$ Tesla. The normal to the plane of loop makes an angle of $60°$ with the field. The maximum induced emf produced in the loop is __________ mV.

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
$12$
B
$18$
C
$21$
D
$24$

Did you get this right?

Solution

Given:

Side of the square loop, $a = 2$ cm $= 2 \times 10^{-2}$ m

Area of the loop, $A = a^2 = 4 \times 10^{-4}$ m $^2$

Magnetic field, $B = 0.4 \sin(300t)$ T

Angle between the normal to the plane and the magnetic field, $\theta = 60^\circ$

The magnetic flux through the loop is given by:

$\Phi = \vec{B} \cdot \vec{A} = B A \cos\theta$

$\Phi = (0.4 \sin(300t))(4 \times 10^{-4}) \cos(60^\circ)$

$\Phi = 1.6 \times 10^{-4} \sin(300t) \times \dfrac{1}{2}$

$\Phi = 0.8 \times 10^{-4} \sin(300t)$ Wb

By Faraday's law of induction, the induced emf is:

$e = -\dfrac{d\Phi}{dt}$

$e = -\dfrac{d}{dt} (0.8 \times 10^{-4} \sin(300t))$

$e = -0.8 \times 10^{-4} \times 300 \cos(300t)$

$e = -240 \times 10^{-4} \cos(300t)$ V

$e = -24 \times 10^{-3} \cos(300t)$ V

The maximum induced emf is the amplitude of $e$ :

$e_{\text{max}} = 24 \times 10^{-3}$ V $= 24$ mV

Answer: $24$

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Solution

Given:

Side of the square loop, $a = 2$ cm $= 2 \times 10^{-2}$ m

Area of the loop, $A = a^2 = 4 \times 10^{-4}$ m $^2$

Magnetic field, $B = 0.4 \sin(300t)$ T

Angle between the normal to the plane and the magnetic field, $\theta = 60^\circ$

The magnetic flux through the loop is given by:

$\Phi = \vec{B} \cdot \vec{A} = B A \cos\theta$

$\Phi = (0.4 \sin(300t))(4 \times 10^{-4}) \cos(60^\circ)$

$\Phi = 1.6 \times 10^{-4} \sin(300t) \times \dfrac{1}{2}$

$\Phi = 0.8 \times 10^{-4} \sin(300t)$ Wb

By Faraday's law of induction, the induced emf is:

$e = -\dfrac{d\Phi}{dt}$

$e = -\dfrac{d}{dt} (0.8 \times 10^{-4} \sin(300t))$

$e = -0.8 \times 10^{-4} \times 300 \cos(300t)$

$e = -240 \times 10^{-4} \cos(300t)$ V

$e = -24 \times 10^{-3} \cos(300t)$ V

The maximum induced emf is the amplitude of $e$ :

$e_{\text{max}} = 24 \times 10^{-3}$ V $= 24$ mV

Answer: $24$

JEE Main 2026 — Physics Electromagnetism

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Solution

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JEE Main 2026 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism