A charged particle of mass ‘m’ and charge ‘q’ moving under the influence of uniform electric field $E \vec{i}$ and a uniform magnetic field $B \vec{k}$ follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, $v \vec{i}$ and $- 2 v \vec{j}$ . Then which of the following statements (A,B,C,D) are the correct ? (Trajectory shown is schematic and not to scale)
A) $E = \frac{3}{4} (\frac{m v^{2}}{q a})$
B) Rate of work done by the electric field at P is $\frac{3}{4} (\frac{m v^{3}}{a})$
C) Rate of work done by both the fields at Q is zero
D) The difference between the magnitude of angular momentum of the particle at P and Q is $2 m a v$

Difficult

Solution

   $\begin{array}{l} W o r k d o n e b y m a g n e t i c f i e l d i s z e r o, b e c a u s e a t a l l p o i n t s f i e l d i s p e r p e n d i c u l a r t o v e l o c i t y . \\ W o r k d o n e b y e l e c t r i c f i e l d, W_{E} = \int_{0}^{2 a} (q E) d x = 2 q E a \\ A p p l y i n g w o r k e n e r g y t h e o r e m, \\ W_{B} + W_{E} = K_{f} - K_{i} \\ \Rightarrow 0 + 2 q E a = \frac{1}{2} m {(2 v)}^{2} - \frac{1}{2} m {(v)}^{2} \\ \Rightarrow 0 + q E (2 a) = \frac{1}{2} m [4 ν^{2} - ν^{2}] \end{array}$
$\Rightarrow E = \frac{3}{4} (\frac{m v^{2}}{q a})$
Rate of work done at P is    $P o w e r = \vec{F} . \vec{v} = (q E \hat{i}) . (v \hat{i}) = q E v = \frac{3}{4} (\frac{m v^{3}}{a})$
Rate of work done at Q Is $P o w e r = \vec{F} . \vec{v} = (q E \hat{i}) . (- 2 v \hat{j}) = q E v = 0$

$\begin{array}{l} A n g u l a r m o m e n t u m a t P, {\vec{L}}_{P} = m \vec{r} \times \vec{v} = m (a \hat{j}) \times v (\hat{i}) \\ \Rightarrow {\vec{L}}_{P} = m v a (- \hat{k}) \\ A n g u l a r m o m e n t u m a t Q, {\vec{L}}_{Q} = m \vec{r} \times \vec{v} = m (2 a \hat{i}) \times 2 v (- {\hat{j}}^{}) \\ \Rightarrow {\vec{L}}_{Q} = 4 m v a (- \hat{k}) \\ ∴ ∣ {\vec{L}}_{Q} ∣ - ∣ {\vec{L}}_{P} ∣ = 4 m v a - m v a = 3 m v a \end{array}$

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