A particle of charge q and mass m is projected from origin with an initial velocity v = (v0/√2 x + v0/√2 y). There exists a uniform magnetic field B = B0 z and a space varying electric field E = E0 e^(-λx) x within the region 0 ≤ x ≤ L. After travelling a distance such that x-coordinate has changed from x=0 to x=L, the change in the kinetic energy is _______.

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q38 – JEE-MAIN 2026

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Home Physics Work, Energy & PowerQuestion #648

A particle of charge

q

and mass

m

is projected from origin with an initial velocity

\vec{v} = (\frac{v_{0}}{\sqrt{2}} \hat{x} + \frac{v_{0}}{\sqrt{2}} \hat{y})

. There exists a uniform magnetic field

\vec{B} = B_{0} \hat{z}

and a space varying electric field

\vec{E} = E_{0} e^{- λ x} \hat{x}

within the region

0 \leq x \leq L

. After travelling a distance such that

x

-coordinate has changed from

x = 0

x = L

, the change in the kinetic energy is _______.

Clue 1 of 3: Work-Energy Theorem

The change in kinetic energy of a particle is equal to the total work done on it by all forces.

🧠 Full Solution Path

Step 1: Apply Work-Energy Theorem and Identify Forces✦ Active

The change in kinetic energy ( $Δ K$ ) is equal to the total work done ( $W_{t o t a l}$ ). The total work is the sum of work done by the electric field ( $W_{E}$ ) and the magnetic field ( $W_{B}$ ). The magnetic force is always perpendicular to the velocity, so the work done by the magnetic field is zero ( $W_{B} = 0$ ). Therefore, the change in kinetic energy is solely due to the work done by the electric field.

Δ K = W_{t o t a l} = W_{E} + W_{B}

{\vec{F}}_{B} = q (\vec{v} \times \vec{B}) ⟹ W_{B} = 0

Δ K = W_{E}

Step 2: Calculate Work Done by Electric Field○ Expand

The electric force on the particle is ${\vec{F}}_{E} = q \vec{E}$ . Given $\vec{E} = E_{0} e^{- λ x} \hat{x}$ , the force is ${\vec{F}}_{E} = q E_{0} e^{- λ x} \hat{x}$ . The particle's x-coordinate changes from $x = 0$ to $x = L$ , so the infinitesimal displacement vector is $d \vec{r} = d x \hat{x}$ . The work done by the electric field is the integral of the dot product of the force and displacement.

W_{E} = \int_{0}^{L} {\vec{F}}_{E} \cdot d \vec{r}

{\vec{F}}_{E} \cdot d \vec{r} = (q E_{0} e^{- λ x} \hat{x}) \cdot (d x \hat{x}) = q E_{0} e^{- λ x} d x

⟹ W_{E} = \int_{0}^{L} q E_{0} e^{- λ x} d x

💡 Teacher's Secret Hint

Remember that only the component of force parallel to displacement contributes to work.

Step 3: Evaluate the Integral○ Expand

Now, we evaluate the definite integral to find the total work done by the electric field, which is equal to the change in kinetic energy.

W_{E} = q E_{0} \int_{0}^{L} e^{- λ x} d x

W_{E} = q E_{0} {[\frac{e^{- λ x}}{- λ}]}_{0}^{L}

W_{E} = q E_{0} (\frac{e^{- λ L}}{- λ} - \frac{e^{- λ (0)}}{- λ})

W_{E} = q E_{0} (\frac{e^{- λ L}}{- λ} - \frac{1}{- λ})

W_{E} = q E_{0} (\frac{1}{λ} - \frac{e^{- λ L}}{λ})

W_{E} = \frac{q E_{0}}{λ} (1 - e^{- λ L})

Thus, the change in kinetic energy is $\frac{q E_{0}}{λ} (1 - e^{- λ L})$ , which corresponds to option 1.

💡 Teacher's Secret Hint

Pay attention to the limits of integration and the sign of the exponent during integration.

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Physics Question 38 – JEE-MAIN 2026

Clue 1 of 3: Work-Energy Theorem

🧠 Full Solution Path