JEE Challenger
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PhyMedMain 2026 • 5th Apr - Shift 2

Change in Kinetic Energy of Charged Particle in Electric and Magnetic Fields

A particle of charge qq and mass mm is projected from origin with an initial velocity v⃗=(v02x^+v02y^)\vec{v} = \left( \frac{v_0}{\sqrt{2}} \hat{x} + \frac{v_0}{\sqrt{2}} \hat{y} \right). There exists a uniform magnetic field B⃗=B0z^\vec{B} = B_0 \hat{z} and a space varying electric field E⃗=E0e−λxx^\vec{E} = E_0 e^{-\lambda x} \hat{x} within the region 0⩽x⩽L0 \leqslant x \leqslant L. After travelling a distance such that xx-coordinate has changed from x=0x=0 to x=Lx=L, the change in the kinetic energy is ______________.

Options

A

qE0λ[1−e−λL]\frac{q E_0}{\lambda} \left[ 1 - e^{-\lambda L} \right]

Correct
B

(v0qB02λ)[2−e−2λL]\left( \frac{v_0 q B_0}{2\lambda} \right) \left[ 2 - e^{-2\lambda L} \right]

C

qE0λ[1+e−λL]\frac{q E_0}{\lambda} \left[ 1 + e^{-\lambda L} \right]

D

q(E0+v0B0λ)[1−e−λL/2]q \left( \frac{E_0 + v_0 B_0}{\lambda} \right) \left[ 1 - e^{-\lambda L/2} \right]

Topics & Concepts

Moving Charges and MagnetismMagnetic ForceWork-Energy TheoremElectric Field

Step-by-Step Solution

To find the change in kinetic energy of the charged particle, we can apply the Work-Energy Theorem, which states that the net work done by all forces acting on a particle is equal to the change in its kinetic energy:

ΔK=Wnet=WE+WB\Delta K = W_{\text{net}} = W_E + W_B

where:

  • WEW_E is the work done by the electric field E⃗\vec{E}.
  • WBW_B is the work done by the magnetic field B⃗\vec{B}.

Step 1: Work done by the Magnetic Field (WBW_B)

The magnetic force acting on a moving charge qq is given by the Lorentz force formula: F⃗B=q(v⃗×B⃗)\vec{F}_B = q(\vec{v} \times \vec{B})

Since the magnetic force F⃗B\vec{F}_B is always perpendicular to the velocity vector v⃗\vec{v} at every instant (F⃗B⋅v⃗=0\vec{F}_B \cdot \vec{v} = 0), the power delivered by the magnetic field is zero. Consequently, the work done by the magnetic field is zero: WB=∫F⃗B⋅dr⃗=0W_B = \int \vec{F}_B \cdot d\vec{r} = 0

Step 2: Work done by the Electric Field (WEW_E)

The electric force acting on the particle is: F⃗E=qE⃗=qE0e−λxx^\vec{F}_E = q \vec{E} = q E_0 e^{-\lambda x} \hat{x}

The infinitesimal work done by this force for a displacement dr⃗=dxx^+dyy^+dzz^d\vec{r} = dx \hat{x} + dy \hat{y} + dz \hat{z} is: dWE=F⃗E⋅dr⃗=(qE0e−λxx^)⋅(dxx^+dyy^+dzz^)=qE0e−λxdxdW_E = \vec{F}_E \cdot d\vec{r} = \left( q E_0 e^{-\lambda x} \hat{x} \right) \cdot \left( dx \hat{x} + dy \hat{y} + dz \hat{z} \right) = q E_0 e^{-\lambda x} dx

Integrating from x=0x = 0 to x=Lx = L: WE=∫0LqE0e−λxdxW_E = \int_{0}^{L} q E_0 e^{-\lambda x} dx

WE=qE0[e−λx−λ]0LW_E = q E_0 \left[ \frac{e^{-\lambda x}}{-\lambda} \right]_{0}^{L}

WE=−qE0λ(e−λL−e0)W_E = -\frac{q E_0}{\lambda} \left( e^{-\lambda L} - e^0 \right)

WE=qE0λ(1−e−λL)W_E = \frac{q E_0}{\lambda} \left( 1 - e^{-\lambda L} \right)

Step 3: Total Change in Kinetic Energy

Substituting WEW_E and WBW_B back into the Work-Energy equation: ΔK=WE+WB=qE0λ(1−e−λL)+0\Delta K = W_E + W_B = \frac{q E_0}{\lambda} \left( 1 - e^{-\lambda L} \right) + 0

ΔK=qE0λ[1−e−λL]\Delta K = \frac{q E_0}{\lambda} \left[ 1 - e^{-\lambda L} \right]

Conclusion

The change in kinetic energy of the particle is qE0λ[1−e−λL]\frac{q E_0}{\lambda} \left[ 1 - e^{-\lambda L} \right], which corresponds to Option A.

Change in Kinetic Energy of Charged Particle in Electric and Magnetic Fields | Physics PYQ Solution - JEE Challenger