A particle of charge per unit mass α is released from the origin with velocity v → = v 0 i ^ in the magnetic field B → = - B 0 k ^ for x ≤ 3 2 v 0 B 0 α and B → = 0 for x > 3 2 v 0 B 0 α The x-coordinate of the particle at time t > π 3 B 0 α would be

r = mv 0 B 0 q = v 0 B 0 α , x r = 3 2 = sin θ ⇒ θ = 6 0 ∘ t OA = T 6 = π 3 B 0 α Therefore x-coordinate of particle at any time t > π 3 B 0 α will be x = 3 2 v 0 B 0 α + v 0 t - π 3 B 0 α cos 6 0 ∘ = 3 2 v 0 B 0 α + v 0 2 t - π 3 B 0 α

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PhysicsMagnetismMotion analysis in electric & magnetic fieldMedium2 minPYQ_2020

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A particle of charge per unit mass $α$ is released from the origin with velocity $\vec{v} = v_{0} \hat{i}$ in the magnetic field
$\vec{B} = - B_{0} \hat{k}$ for $x \leq \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α}$
and $\vec{B} = 0$ for $x > \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α}$
The x-coordinate of the particle at time $t (> \frac{π}{3 B_{0} α})$ would be

Options:

Answer:

Solution:

$r = \frac{{mv}_{0}}{B_{0} q} = \frac{v_{0}}{B_{0} α}, \frac{x}{r} = \frac{\sqrt{3}}{2} = sin θ \Rightarrow θ = 6 0^{\circ}$

$t_{OA} = \frac{T}{6} = \frac{π}{3 B_{0} α}$

Therefore x-coordinate of particle at any time $t > \frac{π}{3 B_{0} α}$ will be

$x = \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α} + v_{0} (t - \frac{π}{3 B_{0} α}) cos 6 0^{\circ}$

$= \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α} + \frac{v_{0}}{2} (t - \frac{π}{3 B_{0} α})$

Stream:JEESubject:PhysicsTopic:MagnetismSubtopic:Motion analysis in electric & magnetic field

⏱ 2mℹ️ Source: PYQ_2020

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