JEE Main 2026 · 5 Apr · Shift 2 · Physics Q38

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PhysicsSingle correct+4 / -1Question 38 of 75

A particle of charge $q$ and mass $m$ is projected from origin with an initial velocity $\vec v=\left(\dfrac{v_0}{\sqrt2}\hat x+\dfrac{v_0}{\sqrt2}\hat y\right)$ . There exists a uniform magnetic field $\vec B=B_0\hat z$ and a space varying electric field $\vec E=E_0e^{-\lambda x}\hat x$ within the region $0\le x\le L$ . After travelling a distance such that $x$ -coordinate has changed from $x=0$ to $x=L$ , the change in the kinetic energy is ________.

\dfrac{qE_0}\lambda\left[1-e^{-\lambda L}\right]

\left(\dfrac{v_0qB_0}{2\lambda}\right)\left[2-e^{-2\lambda L}\right]

\dfrac{qE_0}\lambda\left[1+e^{-\lambda L}\right]

q\left(\dfrac{E_0+v_0B_0}\lambda\right)\left[1-e^{-\lambda L/2}\right]

Answer and worked solution

Official answer: A

Work. The magnetic force does no work, so only the electric field changes the kinetic energy:
$\Delta K=\int_0^LqE_0e^{-\lambda x}\,dx=\frac{qE_0}\lambda\left[1-e^{-\lambda L}\right].$

(A)

Topics:Motion in magnetic and electric fields Work-energy theorem

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JEE Main 2026 · 5 Apr · Shift 2, Physics question 38