Question

Crossed E and B Fields. A particle with initial velocity$\overrightarrow{\mathbf{v}}\mathbf{=}\left(5.85\times {10}^{3}m/s\right)\hat{\mathbf{j}}$enters a region of uniform electric and magnetic fields. The magnetic field in the region is$\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{35}\mathbf{T}\mathbf{)}\mathit{k}$. Calculate the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge (a) +0.640 nC and (b) -0.320 nC. You can ignore the weight of the particle.

Short answer

a) The magnitude and direction of the charged particle of charge +0.640 nC is $\left(7898N/C\right)\hat{\mathbf{i}}$

b) The magnitude and direction of the charged particle of charge -0.320 nC is $\mathbf{(}\mathbf{-}\mathbf{7898}\mathbf{N}\mathbf{/}\mathbf{C}\mathbf{)}\hat{\mathbf{i}}$

Step-by-step solution

Magnetic force and electric force

The magnetic force on a particle is given by

$\overrightarrow{{\mathbf{F}}_{\mathbf{B}}}\mathbf{=}\mathbf{q}\mathbf{(}\overrightarrow{\mathbf{v}}\mathbf{\times }\overrightarrow{\mathbf{B}}\mathbf{)}$

Here q is the charge of the particle and v is the velocity of the particle

And the electric force is

$\overrightarrow{{\mathbf{F}}_{\mathbf{E}}}\mathbf{=}\mathbf{q}\overrightarrow{\mathbf{E}}$

Determine the magnitude and direction of the first particle

(a)

The magnetic force is

$\overrightarrow{{\mathbf{F}}_{\mathbf{B}}}\mathbf{=}\mathbf{q}\mathbf{(}\overrightarrow{\mathbf{v}}\mathbf{\times }\overrightarrow{\mathbf{B}}\mathbf{)}$

And the electric force on the particle is

$\overrightarrow{{\mathbf{F}}_{\mathbf{E}}}\mathbf{=}\mathbf{q}\overrightarrow{\mathbf{E}}$

Both forces are equal

Therefore,

$\begin{array}{r}\overrightarrow{{\mathbf{F}}_{\mathbf{E}}}\mathbf{=}\overrightarrow{{\mathbf{F}}_{\mathbf{B}}}\\ \mathbf{qvB}\mathbf{=}\mathbf{qE}\\ \mathbf{E}\mathbf{=}\mathbf{vB}\end{array}$

Put the values in the equation

$\begin{array}{r}\mathbf{E}\mathbf{=}\left(5.85\times {10}^{3}m/s\right)\mathbf{(}\mathbf{1}\mathbf{.35}\mathbf{T}\mathbf{)}\\ \mathbf{=}\mathbf{7898}\mathbf{N}\mathbf{/}\mathbf{C}\end{array}$

And the direction is $\hat{\mathbf{j}}\mathbf{\times }\mathbf{(}\mathbf{-}\mathbf{k}\mathbf{)}\mathbf{=}\mathbf{-}\hat{\mathbf{i}}$

Therefore, the magnitude and direction of the charged particle of charge +0.640 nC is$\left(7898N/C\right)\hat{\mathbf{i}}$

Determine the direction and magnitude of the second particle

(b)

Same as the previous one,

The electric field of the second particle is

$\begin{array}{r}\overrightarrow{\mathbf{E}}\mathbf{=}\mathbf{vB}\mathbf{(}\mathbf{-}\widehat{\mathbf{i}}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{E}\mathbf{=}\left(5.85\times {10}^{3}m/s\right)\mathbf{(}\mathbf{1}\mathbf{.35}\mathbf{T}\mathbf{)}\mathbf{(}\mathbf{-}\widehat{\mathbf{i}}\mathbf{)}\\ \mathbf{=}\mathbf{-}\mathbf{7898}\mathbf{N}\mathbf{/}\mathbf{C}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\end{array}$

Therefore, the magnitude and direction of the charged particle of charge +0.640 nC is$\left(-7898N/C\right)\hat{\mathbf{i}}$