Question

What is the force (magnitude and direction) on the proton in FIGURE P31.29? Give the direction as an angle $\text{CW}$or $\text{cCW}$from vertical.

Short answer

The force, $F_B=1.6\times {10}^{-13}\mathrm{N}$

The force, $F_e=1.6\times {10}^{-13}\mathrm{N}$

The net force,$F_{net}=2.26\times {10}^{-13}\mathrm{N}$

Step-by-step solution

Find the direction of electrostatic force.

With in presence of an electric field, there is an electrostatic force called $F_E$.

Find FB, Fe, Fnet 

The magnetic force flows north and use the right - hand principle.

Calculate $F_B$

localid="1648970948872" $\begin{array}{r}{F}_B=qvB\end{array}$

$F_B=1.6\times {10}^{-19}\times {10}^7\times 0.1$ ( enter the quantities in the formula)

$F_B=1.6\times {10}^{-13}\mathrm{N}$

Calculate localid="1648971852105" $F_e$

localid="1648971000005" $F_e=eE$

$F_e=1.6\times {10}^{-19}\times {10}^{-6}$ ( enter the quantities in the formula)

$F_e=1.6\times {10}^{-13}\mathrm{N}$

Calculate $F_{net}$

localid="1648971062338" $F_{net}=\sqrt{F_B^2+F_e^2}$

$F_{net}=\sqrt{2}\times F_B$

$F_{net}=\sqrt{2}\times 1.6\times {10}^{-13}$ ( enter the quantities in the formula)

$F_{net}=2.26\times {10}^{-13}\mathrm{N}$

Find direction and θ

To find the direction and$\theta$

$\tan \theta =\frac{F_B}{F_e}$

$\tan \theta =\frac{1.6\times {10}^{-13}}{1.6\times {10}^{-13}}$ ( enter the quantities in the formula)

$\tan \theta =1\Rightarrow \theta ={45}^{\circ }$