JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A particle of charge \(q\) and mass \(m\) is projected from origin with an initial velocity \(\vec{v} = \left(\dfrac{v_0}{\sqrt{2}}\hat{x} + \dfrac{v_0}{\sqrt{2}}\hat{y}\right)\). There exists a uniform magnetic field \(\vec{B} = B_0 \hat{z}\) and a space varying electric field \(\vec{E} = E_0 e^{-\lambda x}\hat{x}\) within the region \(0 \leq x \leq L\). After travelling a distance such that \(x\)-coordinate has changed from \(x=0\) to \(x=L\), the change in the kinetic energy is _______.
- A \(\dfrac{qE_0}{\lambda}[1 - e^{-\lambda L}]\)
- B \(\left(\dfrac{v_0 q B_0}{2\lambda}\right)[2 - e^{-2\lambda L}]\)
- C \(\dfrac{qE_0}{\lambda}[1 + e^{-\lambda L}]\)
- D \(q\left(\dfrac{E_0 + v_0 B_0}{\lambda}\right)[1 - e^{-\lambda L/2}]\)
Answer & Solution
Correct Answer
(A) \(\dfrac{qE_0}{\lambda}[1 - e^{-\lambda L}]\)
Step-by-step Solution
Detailed explanation
According to the work-energy theorem, the change in kinetic energy of a particle is equal to the total work done by all the forces acting on it. The forces acting on the particle are the electric force and the magnetic force. The magnetic force is given by…
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