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Physics Question 39 – JEE-MAIN 2026

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A square loop of side 2 cm is placed in a time varying magnetic field with magnitude as B=0.4sin⁡(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.

The induced electromotive force (EMF) in a loop is proportional to the rate of change of magnetic flux through the loop.

Step 1: Calculate the area of the loop and magnetic flux✦ Active

The side of the square loop is s=2 cm=0.02 m. The area of the loop is A=s2=(0.02)2=4×10−4 m2. The magnetic field is B=0.4sin⁡(300t) T. The angle between the normal to the loop and the field is θ=60∘. The magnetic flux is given by:

ΦB=BAcos⁡θ

Substitute the values:

ΦB=(0.4sin⁡(300t))(4×10−4)cos⁡(60∘) ΦB=(0.4sin⁡(300t))(4×10−4)(0.5) ΦB=8×10−5sin⁡(300t) Wb
Step 2: Calculate the induced EMF○ Expand

According to Faraday's Law, the induced EMF is ε=−dΦBdt.

ε=−ddt(8×10−5sin⁡(300t)) ε=−8×10−5(300cos⁡(300t)) ε=−2400×10−5cos⁡(300t) ε=−0.024cos⁡(300t) V
💡 Teacher's Secret Hint

Remember to apply the chain rule when differentiating sin⁡(300t).

Step 3: Determine the maximum induced EMF and convert units○ Expand

The maximum value of cos⁡(300t) is 1. Therefore, the maximum induced EMF is:

εmax=|−0.024×1|=0.024 V

Converting to millivolts:

εmax=0.024×1000 mV=24 mV
💡 Teacher's Secret Hint

The maximum value of a sinusoidal function (like cos⁡(x)) is 1. The negative sign in Faraday's law indicates the direction of induced current, not the magnitude of EMF.

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