First slide

Force on a charged particle moving in a magnetic field

Question

A particle of charge + q and mass m moving under the influence of a uniform electric field E i and a uniform magnetic field B k follows trajectory from P to Q as shown in fig. The velocities at P and Q are v i and
- 2 v j respectively. Which of the following statements (s) is/are correct

Difficult

A

$E = \frac{3}{4} \frac{{mv}^{2}}{qa}$
B

rate of work done by electric field at P is $\frac{3}{4} \frac{{mv}^{2}}{qa}$
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

Solution

(i) $W = (K . E .)_{Q} - (K . E .)_{P}$
$∴ qE \times 2 a = \frac{1}{2} m (2 v)^{2} - \frac{1}{2} {mv}^{2}$
or $qE \times 2 a = \frac{1}{2} m \times 3 v^{2}$
$E = \frac{3}{4} (\frac{{mv}^{2}}{qa})$
(ii) The rate of work done by electric field at P
$\begin{array}{l} = Fv = qEv \\ = q [\frac{3}{4} (\frac{{mv}^{2}}{qa}) v] = \frac{3}{4} (\frac{{mv}^{3}}{a}) \end{array}$
(iii) At Q, the force due to electric field is q E along X-axis. Here velocity v is along negative y-axis. Hence rate of doing work F_e .v= 0 $(∵ θ = 90^{\circ})$
The force due to magnetic field
F_m= q (v x B)
Here F_m- is also perpendicular to velocity vector v,
$∴$ Rate of doing work in magnetic field
F_m. v = 0

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