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Electric Field

In a certain ...

Question

In a certain region, uniform electric field $\to E = - E_{0}^k$ and uniform magnetic field $\to B = - B_{0}^k$ are present. At time $t = 0$ , a particle of mass $m$ and charge $q$ is given a velocity $v = v_{0}^j + v_{0}^k$ . Find the time when the speed of the particle is minimum.

\frac{4 m v_{0}}{q E_{0}}

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\frac{3 m v_{0}}{q E_{0}}

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\frac{2 m v_{0}}{q E_{0}}

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\frac{m v_{0}}{q E_{0}}

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Solution

The correct option is D $\frac{m v_{0}}{q E_{0}}$
Given,
Electric field $\to E = - E_{0}^k$ ,
Magnetic field $\to B = - B_{0}^k$ ,
Velocity $v = v_{0}^j + v_{0}^k$ .

The acceleration of the particle,

$_{E} = \frac{q E_{0}}{m}$ (along $- v e$ z-direction)

Electric field will make z-component of velocity zero.

At that time, the speed of the particle will be minimum and that minimum speed is due to the other component ,i.e, y-component.

This is minimum when,

$v_{z} = u_{z} + a_{z} t$

$\Rightarrow 0 = v_{0} - \frac{q E_{0}}{m} t$

$∴ t = \frac{m v_{0}}{q E_{0}}$

Hence, option (d) is the correct answer.

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Electric Field

Standard XII Physics

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In a certain region, uniform electric field →E=−E0^k and uniform magnetic field →B=−B0^k are present. At time t=0, a particle of mass m and charge q is given a velocity v=v0^j+v0^k. Find the time when the speed of the particle is minimum.

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