a point charge of 5.7 μc moves at 4.5×10^5 m/s in a magnetic field that has a field strength of 3.2 mt, as…

Explanation:

Step1: Identify the 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.

Step2: Convert units

Convert the charge $q = 5.7\ \mu C=5.7\times10^{- 6}\ C$, the magnetic - field strength $B = 3.2\ mT = 3.2\times10^{-3}\ T$, and $\theta = 37^{\circ}$, $v = 4.5\times10^{5}\ m/s$.

Step3: Substitute values into the formula

$F=(5.7\times10^{-6}\ C)\times(4.5\times10^{5}\ m/s)\times(3.2\times10^{-3}\ T)\times\sin37^{\circ}$. We know that $\sin37^{\circ}\approx0.6$. $F=(5.7\times10^{-6})\times(4.5\times10^{5})\times(3.2\times10^{-3})\times0.6$. First, calculate $(5.7\times10^{-6})\times(4.5\times10^{5}) = 5.7\times4.5\times10^{-6 + 5}=25.65\times10^{-1}=2.565$. Then, $2.565\times(3.2\times10^{-3})\times0.6=(2.565\times3.2\times0.6)\times10^{-3}$. $2.565\times3.2 = 8.208$, and $8.208\times0.6 = 4.9248$. So, $F = 4.9248\times10^{-3}\ N\approx4.9\times10^{-3}\ N$.

Answer:

$4.9\times10^{-3}\ N$