An electron with mass 'm' with an initial velocity (t=0) v = v0 i (v0 > 0) enters a magnetic field B = B0 j. If the initial de-Broglie wavelength at t=0 is λ0 then its value after time 't' would be :

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

Physics Q33 – JEE-MAIN 2025

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Home Physics Moving ChargesQuestion #22

An electron with mass '

m

' with an initial velocity (

t = 0

)

\vec{v} = v_{0} \hat{i}

(

v_{0} > 0

) enters a magnetic field

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

. If the initial de-Broglie wavelength at

t = 0

λ_{0}

then its value after time '

t

' would be :

Clue 1 of 3: Magnetic Force on a Charged Particle

Recall that the magnetic force on a charged particle is always perpendicular to its velocity.

🥷

Ninja Strategy•Conservation of Speed

Recognize that a magnetic force does no work on a charged particle, thus its speed and kinetic energy remain constant, leading to a constant de-Broglie wavelength.

🧠 Full Solution Path

Step 1: Analyze the effect of magnetic force on speed✦ Active

The magnetic force $\vec{F} = q (\vec{v} \times \vec{B})$ is always perpendicular to the velocity vector $\vec{v}$ . Therefore, the work done by the magnetic force, $W = \int \vec{F} \cdot d \vec{r} = \int \vec{F} \cdot (\vec{v} d t)$ , is always zero because $\vec{F} \cdot \vec{v} = 0$ . Since no work is done, the kinetic energy of the electron, $K = \frac{1}{2} m v^{2}$ , remains constant.

K (t) = K (0) ⟹ \frac{1}{2} m v (t)^{2} = \frac{1}{2} m v_{0}^{2}

This implies that the speed of the electron, $v (t)$ , remains constant and equal to its initial speed $v_{0}$ .

Step 2: Relate speed to de-Broglie wavelength○ Expand

The de-Broglie wavelength is given by $λ = \frac{h}{p}$ , where $h$ is Planck's constant and $p$ is the momentum of the particle. The momentum is $p = m v$ . Since the speed $v$ remains constant ( $v (t) = v_{0}$ ), the momentum $p (t) = m v (t) = m v_{0}$ also remains constant.

Step 3: Determine the de-Broglie wavelength at time

t

○ Expand

As the momentum remains constant, the de-Broglie wavelength at any time $t$ will be the same as the initial de-Broglie wavelength $λ_{0}$ .

λ (t) = \frac{h}{m v (t)} = \frac{h}{m v_{0}} = λ_{0}

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Physics Question 33 – JEE-MAIN 2025

Clue 1 of 3: Magnetic Force on a Charged Particle

🧠 Full Solution Path