a point charge of 6.8 μc moves at 6.5×10⁴ m/s at an angle of 15° to a magnetic field that has a field…

Explanation:

Step1: Convert charge to SI units

$q = 6.8\ \mu C=6.8\times 10^{- 6}\ C$

Step2: Recall magnetic - force formula

The formula for the magnetic force on a moving charge is $F = qvB\sin\theta$, where $q$ is the charge, $v$ is the velocity, $B$ is the magnetic - field strength, and $\theta$ is the angle between the velocity and the magnetic field.

Step3: Substitute values into the formula

$v = 6.5\times 10^{4}\ m/s$, $B = 1.4\ T$, $\theta = 15^{\circ}$, and $q = 6.8\times 10^{-6}\ C$. $F=(6.8\times 10^{-6}\ C)\times(6.5\times 10^{4}\ m/s)\times(1.4\ T)\times\sin(15^{\circ})$ We know that $\sin(15^{\circ})=\sin(45^{\circ}- 30^{\circ})=\sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.259$ $F=(6.8\times 10^{-6})\times(6.5\times 10^{4})\times(1.4)\times0.259$ $F=(6.8\times6.5\times1.4\times0.259)\times10^{-6 + 4}$ $F=(6.8\times6.5\times1.4\times0.259)\times10^{-2}$ $6.8\times6.5 = 44.2$, $44.2\times1.4 = 61.88$, $61.88\times0.259\approx16$ $F = 16\times10^{-2}\ N=1.6\times 10^{-1}\ N$

Answer:

A. $1.6\times 10^{-1}\ N$