Question

A particle with charge +q and mass m₁, moving under the influence of a uniform electric field E$\overset{\hat{}}{i}$ and a uniform magnetic field B$\overset{\hat{}}{k}$, follows a trajectory from P to Q as shown. The velocities at P and Q are V$\overset{\hat{}}{i}$ and - 2V$\overset{\hat{}}{j}$. Choose the wrong statement.

• A. E= $\frac { 3 } { 4 }$ $\left( \frac{mv^{2}}{qa} \right)$
• B. The rate of work done by the electric field at P is $\frac { 3 } { 4 }$ $\left( \frac{mv^{3}}{a} \right)$
• C. The rate of work done by the electric field at P is 0.
• D. The rate of work done by both the fields at Q is 0

Answer 1

Answer: (C)

The rate of work done by the electric field at P is 0.

Answer 2

(A),(B),(D), in going from P to Q increase in kinetic energy

= $\frac{1}{2}$m$(2V)^{2} -$ $\frac{1}{2}$m$V^{2}$= $\frac{1}{2}$m$\left( 3v^{2} \right)^{}$= work done by electric field.

or $\frac{3}{2}$m$v^{2}$ = Eq x 2a or E=$\frac{3}{4}\left( \frac{mv^{2}}{qa} \right)$

The rate of work done by E at P = force due to E x velocity.

= (qE)v = qv $\left\lbrack \frac{3}{4}\left( \frac{mv^{2}}{qa} \right) \right\rbrack$=$\frac{3}{4}\left( \frac{mv^{3}}{a} \right)$

At q,$\bar{V}$ is perpendicular to $\bar{E}$ and $\bar{B}$, and no work is done by either field