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ser-zykov [4K]
1 year ago
8

A 1.25 x 10-4 C charge is moving5200 m/s at 37.0° to a magnetic fieldof 8.49 x 10-4 T. What is the magneticforce on the charge?

Physics
1 answer:
juin [17]1 year ago
7 0

0.0003321 Newtons

Explanation

We are given the charge, its velocity, and the magnetic field strength and direction. We can thus use the equation to find the force

\begin{gathered} F=qvBsinθ \\ where\text{ F is the magnetic force} \\ q\text{ is the charge} \\ v\text{ is the velocity of the charge} \\ Bis\text{ the magnetic field} \\ \theta\text{ is the angle} \end{gathered}

so

Step 1

Let

\begin{gathered} q=1.25*10^{-4\text{ }}C \\ v=5200\frac{m}{s} \\ \theta=37\text{ \degree} \\ B=8.49*10^{-4}T \end{gathered}

now, replace

\begin{gathered} F=qvBs\imaginaryI n\theta \\ F=1.25*10^{-4}\text{ C*5200 }\frac{m}{s}*8.49*10^{-4}Tsin(37) \\ F=0.0003321\text{ Newtons} \\  \end{gathered}

so, the answer is

0.0003321 Newtons

I hope this helps you

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