First slide

Force on a charged particle moving in a magnetic field

Question

A charged particle enters a uniform magnetic field with velocity v₀ perpendicular to it. The length of magnetic field is $x = \frac{\sqrt{3}}{2} R,$ where R is the radius of the circular path of the particle in the field. The magnitude of change in velocity of the particle when it comes out of the field is

Moderate

A

$2 v_{0}$
B

$v_{0} / 2$
C

$\frac{\sqrt{3} v_{0}}{2}$
D

$v_{0}$

Solution

The particle will come out of the magnetic field at an angle $θ = 60^{0}$ with the original direction.
$\begin{array}{l} Δ \vec{v} = (v_{0} \cos 60 \hat{i} + v_{0} \sin 60 \hat{j} - v_{0} \hat{i}) \\ \Rightarrow |Δ \vec{v}| = v_{0} \end{array}$

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