Question

How fast must a 8.13C charge be moving perpendicular to a 2.61T magnetic field to experience a force of 2.78N?0.131m/sO1.12m/s7.63m/s0.982m/s

Answer 1

0.131m/s

Explanation

$F_B=\lvert q\rvert vBsin\theta$

where

$\begin{gathered} F_Bis\text{ the charge} \\ v\text{ is the velocity} \\ q\text{ is the charge } \\ B\text{ is the magnetil field} \\ \theta\text{ is the angle formed} \end{gathered}$

Step 1

A) let

$\begin{gathered} Charge=q=8.13\text{ C} \\ B=2.61\text{ T} \\ F=2.78\text{ N} \end{gathered}$

B) replace in the formula and solve for v

$\begin{gathered} F_B=\lvert q\rvert vBsin\theta \\ 2.78\text{ N=}\lvert{8.13\text{ C}}\rvert2.61*v \\ 2.78\text{ N}=21.2193*v\text{ } \\ divide\text{ both sides by 21.2193} \\ \frac{2.78N}{21.2193}=\frac{21.2193v}{21.2193} \\ 0.131\frac{m}{s}=v \end{gathered}$

therefore,the answer is 0.131m/s

I hope