Question #21d0f

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Question #21d0f

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Answer 1 · 2016-01-30T01:00:53

Only possible if velocity of the charged particle is $= 1.907 \times 10^{7} m / s$ $=1.907times 10^7 m//s$ . Direction of $\to v$ $vec v$ must be as assumed in the solution.
It is independent of $\frac{e}{m}$ $e/m$

Explanation:

This is a question related to Motion of a Charged Particle in Electric and Magnetic Field. Let us recall the applicable Lorentz Force equation

$\to F = q [\to E + (\to v \times \to B)]$ $vec F=q[vecE + (vecvxxvecB)]$
where $q$ $q$ is the charge of the particle, $\to E$ $vecE$ is the electric field, $\to v$ $vecv$ is the velocity of the charged particle and $\to B$ $vecB$ is the magnetic field.

Assumption of direction of three Vectors

Let the charged particle move in the direction given by $ˆ x$ $hat x$ , the electric field be in the $ˆ y$ $haty$ and magnetic field be in the $ˆ z$ $hatz$ direction.

Electric field $\to E$ $vecE$ will exert force on the charge in its direction, assuming the charge $q$ $q$ to be positive.

The direction of the force exerted by the magnetic field is cross product of velocity and magnetic field vectors and is

$ˆ x \times ˆ z = - ˆ y$ $hat x times hat z=-haty$ .
It is acting in a direction opposite to the electric field vector.
As it is required that the charge should pass through in a straight line, it implies that

Lorentz force must be zero.

$0 = q [∣ ∣ ∣ \to E ∣ ∣ ∣ ˆ y - ∣ ∣ \to v ∣ ∣ . ∣ ∣ ∣ \to B ∣ ∣ ∣ ˆ y]$ $0=q[|vecE|hat y - |vecv|.|vecB|haty]$
$\Rightarrow ∣ ∣ ∣ \to E ∣ ∣ ∣ = ∣ ∣ \to v ∣ ∣ . ∣ ∣ ∣ \to B ∣ ∣ ∣$ $=>|vecE|=|vecv|.|vecB|$
$16728 = 0.000877 \times ∣ ∣ \to v ∣ ∣$ $16728=0.000877times |vecv|$
$∣ ∣ \to v ∣ ∣ \approx 1.907 \times 10^{7} m / s$ $|vecv|approx1.907times 10^7 m//s$

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Question #21d0f

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