Question

a particle of mass 10 g and charge 80 μc moves through a uniform magnetic field, in a region where the free - fall acceleration is 9.8ĵ m/s². the velocity of the particle is a constant 20î km/s, which is perpendicular to the magnetic field. what, then, is the magnetic field? attempts: 0 of 3 used save for later etextbook and media submit answer

Explanation:

Step1: Recall the formula for magnetic - force

The magnetic force on a charged particle is given by $\vec{F}=q\vec{v}\times\vec{B}$, and according to Newton's second - law $\vec{F} = m\vec{a}$. Since $\vec{v}$ is perpendicular to $\vec{B}$, $|\vec{F}|=qvB$ and $|\vec{F}| = ma$. So, $ma=qvB$.

Step2: Rearrange the formula to solve for $B$

We can re - arrange the equation $ma = qvB$ to get $B=\frac{ma}{qv}$.

Step3: Convert the given values to SI units

The mass $m = 10\ g=0.01\ kg$, the charge $q = 80\ \mu C=80\times10^{- 6}\ C$, the velocity $v = 20\ km/s = 20000\ m/s$, and the acceleration $a = 9.8\ m/s^{2}$.

Step4: Substitute the values into the formula

Substitute $m = 0.01\ kg$, $a = 9.8\ m/s^{2}$, $q = 80\times10^{-6}\ C$, and $v = 20000\ m/s$ into $B=\frac{ma}{qv}$.
\[

$$\begin{align*} B&=\frac{0.01\ kg\times9.8\ m/s^{2}}{80\times10^{-6}\ C\times20000\ m/s}\\ &=\frac{0.098}{1.6}\\ &=0.06125\ T \end{align*}$$

\]

Answer:

$0.06125\ T$