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Physics Question 48 – JEE-MAIN 2026

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1 μC charge moving with velocity v→=(i^−2j^+3k^) m/s in the region of magnetic field B→=(2i^+3j^−5k^) T. The magnitude of force acting on it is α×10−6 N. The value of α is ______.

The magnetic force on a moving charge in a magnetic field is given by the Lorentz force law.

Step 1: Calculate the cross product of velocity and magnetic field✦ Active

Given velocity v→=(i^−2j^+3k^) m/s and magnetic field B→=(2i^+3j^−5k^) T. The cross product v→×B→ is calculated as:

v→×B→=|i^j^k^1−2323−5|=i^((−2)(−5)−(3)(3))−j^((1)(−5)−(3)(2))+k^((1)(3)−(−2)(2))

Simplifying the terms:

v→×B→=i^(10−9)−j^(−5−6)+k^(3−(−4))=i^+11j^+7k^
Step 2: Calculate the magnetic force and its magnitude○ Expand

The charge is q=1μC=1×10−6 C. The magnetic force is F→=q(v→×B→).

F→=(1×10−6)(i^+11j^+7k^) N

The magnitude of the force is:

|F→|=(1×10−6)(1)2+(11)2+(7)2=(1×10−6)1+121+49

This simplifies to:

|F→|=(1×10−6)171 N
💡 Teacher's Secret Hint

Ensure correct calculation of the magnitude of the vector.

Step 3: Determine the value of α○ Expand

We are given that the magnitude of the force is |F→|=α×10−6 N. Equating this with our calculated magnitude:

α×10−6=171×10−6

Comparing both sides, we find:

α=171⟹α=171
💡 Teacher's Secret Hint

Pay attention to the units and powers of 10 when comparing the magnitudes.

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