a force of 1.5×10² n is exerted on a charge of 1.4×10⁻⁷ c that is traveling at an angle of 75° to a magnetic…

Explanation:

Step1: Recall the formula for magnetic - force

The formula for the magnetic force on a charged particle is $F = qvB\sin\theta$, where $F$ is the force, $q$ 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. We need to solve for $B$, so we can re - arrange the formula to $B=\frac{F}{qv\sin\theta}$.

Step2: Identify the given values

We are given that $F = 1.5\times10^{2}\ N$, $q = 1.4\times10^{-7}\ C$, $v = 1.3\times10^{6}\ m/s$, and $\theta = 75^{\circ}$. The value of $\sin75^{\circ}\approx0.966$.

Step3: Substitute the values into the formula

$B=\frac{1.5\times10^{2}}{(1.4\times10^{-7})\times(1.3\times10^{6})\times0.966}$. First, calculate the denominator: $(1.4\times10^{-7})\times(1.3\times10^{6})\times0.966=(1.4\times1.3\times0.966)\times10^{-7 + 6}\approx1.76\times10^{-1}$. Then, $B=\frac{1.5\times10^{2}}{1.76\times10^{-1}}=\frac{1.5\times10^{2}}{0.176}\approx8.5\times10^{2}\ T$.

Answer:

$8.5\times10^{2}\ T$ (Second option)