Chapter 28: Q81P (page 834)

A 5.0 $μ C$ particle moves through a region containing the uniform magnetic field localid="1664172266088" $- 20 \overset{⏞}{i} mT$ and the uniform electric field 300j^ V/m. At a certain instant the velocity of the particle is localid="1664172275100" $(17 \overset{⏞}{i} - 11 \overset{⏞}{j} + 7.0 \overset{⏞}{k}) k m / s$ . At that instant and in unit-vector notation, what is the net electromagnetic force (the sum of the electric and magnetic forces) on the particle?

Short Answer

Expert verified

The net electromagnetic force on the particle is $\overset{⇀}{F} = (800 \overset{⏞}{j} - 1100 \overset{⏞}{k}) x 10^{- 6} N$

Step by step solution

Given

$k \overset{⇀}{B} = - 20 \hat{i} m T \overset{⇀}{E} = 300 \hat{j} V / m q = 5 \times 10^{- 6} C$

Understanding the concept

Magnetic force is written from equation 28-3 as a vector product of velocity and magnetic field, and the electric force is the charge times the electric field. So the net force is the addition of magnetic force and electric force.

Formula:

$F ⃗ = q E ⃗ + q (v ⃗ \times B ⃗)$

Calculate the net electromagnetic force on the particle

According to net force formula, we can write

$\begin{array}{l} \vec{F} = q \vec{E} + q (\vec{v} \times \vec{B}) \\ \vec{F} = q [\vec{E} + (\vec{v} \times \vec{B})] \end{array}$

By substituting the value, we can get

localid="1662733743521" $\vec{F} = q [300 \hat{j} + ((17 \hat{i} - 11 \hat{j} + 7 \hat{k}) \times - 20 \hat{i})] \vec{F} = q [300 \hat{j} + (0 \hat{i} - 140 \hat{j} - 220 \hat{k})] \vec{F} = q [160 \hat{j} + - 220 \hat{k}] \vec{F} = 5 \times 10^{- 6} [160 \hat{j} + - 220 \hat{k}] \vec{F} = [800 \hat{j} + - 1100 \hat{k}] \times 10^{- 6} N$