Q.38. A particle of charge q and mass m is projected from origin with an initial velocity . There ex

Q: What is the correct answer for JEE Main Physics Q38?

The correct answer is A. The change in kinetic energy is equal to the work done by the net force. The magnetic force does no work because it is always perpendicular to the velocity. So, we only need to consider the work done by the electric field. Work done by electric field $W_E = \int \vec{F}_E \cdot d\vec{r} = \int (q\vec{E}) \cdot d\vec{r}$. Given $\vec{E} = E_0 e^{-\lambda x} \hat{i}$ and the particle moves from $x=0

Q: Q.38. A particle of charge q and mass m is projected from origin with an initial velocity . There exists a uniform magnetic field and a space varying electric field within the region . After…

The correct answer is A. The change in kinetic energy is equal to the work done by the net force. The magnetic force does no work because it is always perpendicular to the velocity. So, we only need to consider the work done by the electric field. Work done by electric field $W_E = \int \vec{F}_E \cdot d\vec{r} = \int (q\vec{E}) \cdot d\vec{r}$. Given $\vec{E} = E_0 e^{-\lambda x} \hat{i}$ and the particle moves from $x=0

Q: What are the options for this JEE Main Physics question?

A. $\frac{qE_0}{\lambda} [1-e^{-\lambda L}]$ · B. $\frac{qE_0}{2\lambda} [2-e^{-\lambda L}]$ · C. $\frac{qE_0}{\lambda} [1+e^{-\lambda L}]$ · D. $q[E_0 + V_0 B_0][1-e^{-\lambda L}]$

Question

Q.38. A particle of charge q and mass m is projected from origin with an initial velocity . There exists a uniform magnetic field  and a space varying electric field  within the region . After travelling a distance such that x-coordinate has changed from  to , the change in the kinetic energy is .

FirstInTest · Accepted Answer

The change in kinetic energy is equal to the work done by the net force. The magnetic force does no work because it is always perpendicular to the velocity. So, we only need to consider the work done by the electric field. Work done by electric field $W_E = \int \vec{F}_E \cdot d\vec{r} = \int (q\vec{E}) \cdot d\vec{r}$. Given $\vec{E} = E_0 e^{-\lambda x} \hat{i}$ and the particle moves from $x=0$ to $x=L$, $d\vec{r} = dx \hat{i} + dy \hat{j} + dz \hat{k}$. So, $W_E = \int_0^L q (E_0 e^{-\lambda x} \hat{i}) \cdot (dx \hat{i}) = q E_0 \int_0^L e^{-\lambda x} dx$. $W_E = q E_0 \left[ -\frac{1}{\lambda} e^{-\lambda x} \right]_0^L = q E_0 \left( -\frac{1}{\lambda} e^{-\lambda L} - \left( -\frac{1}{\lambda} e^0 \right) \right) = q E_0 \left( \frac{1}{\lambda} - \frac{1}{\lambda} e^{-\lambda L} \right) = \frac{q E_0}{\lambda} (1 - e^{-\lambda L})$. This is the change in kinetic energy. This matches option (1).

Answer

$\frac{qE_0}{\lambda} [1-e^{-\lambda L}]$

Answer

$\frac{qE_0}{2\lambda} [2-e^{-\lambda L}]$

Answer

$\frac{qE_0}{\lambda} [1+e^{-\lambda L}]$

Answer

$q[E_0 + V_0 B_0][1-e^{-\lambda L}]$

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