Magnetic Effects Of Current Question 141
Question: A particle of charge +q and mass m moving under the influence of a uniform electric field $ E\hat{i} $ and a uniform magnetic field $ B\hat{k} $ follows trajectory from P to Q as shown in figure. The velocities at P and Q are $ v\hat{i} $ and $ -2v\hat{j} $ respectively. Which of the following statement(s) is/are correct [IIT 1991; BVP 2003]
Options:
A) $ E=\frac{3}{4}\frac{mv^{2}}{qa} $
B) Rate of work done by electric field at P is $ \frac{3}{4}\frac{mv^{3}}{a} $
C) Rate of work done by electric field at P is zero
D) Rate of work done by both the fields at Q is zero
Show Answer
Answer:
Correct Answer: B
Solution:
Kinetic energy of the particle at point $ P=\frac{1}{2}mv^{2} $ K.E. of the particle at point $ Q=\frac{1}{2}m{{(2v)}^{2}} $ Increase in K.E. $ =\frac{3}{2}mv^{2} $ It comes from the work done by the electric force qE on the particle as it covers a distance 2a along the x-axis. Thus $ \frac{3}{2}mv^{2}=qE\times 2a\Rightarrow E=\frac{3}{4}\frac{mv^{2}}{qa} $ . The rate of work done by the electric field at P $ =F\times v=qE\times v=3\frac{mv^{3}}{4a} $ At $ Q,\ \overrightarrow{F_{e}}=q\overrightarrow{E} $ is along x-axis while velocity is along negative y-axis. Hence rate of work done by electric field $ =\overrightarrow{F_{e}.},\overrightarrow{v,}=0\ \ (\because \theta =90^{o}) $ Similarly, according to equation $ \overrightarrow{F_{m}}=q(\overrightarrow{v,}\times \overrightarrow{B}) $ Force $ \overrightarrow{{F_{\text{m}}}} $ is also perpendicular to velocity vector $ v $ . Hence the rate of work done by the magnetic field = 0
BETA
