In Fig. 28-36, a particle moves along a circle in a region of uniform magnetic field of magnitude$\mathbf{}\mathbf{B}\mathbf{\text{=4.00mT}}$. The particle is either a proton or an electron (you must decide which). It experiences a magnetic force of magnitude ${\mathbf{\text{3.20×10}}}^{\mathbf{\text{-15}}}\mathbf{\text{N}}$. What are (a) the particle’s speed, (b) the radius of the circle, and (c)the period of the motion?

The magnetic field is$\text{B}=\text{4.00mT}\left(\frac{{10}^{-3}\text{\hspace{0.17em}T}}{1mT}\right)=4.00\times {10}^{-3}T$.

Magnetic force is${\text{F}}_{\text{B}}=\text{3.20}\times {\text{10}}^{\text{-15}}\text{\hspace{0.17em}N}$.

Figure 28-36 is the particle in the magnetic field.

Use the concept of magnetic force, centripetal force, and period. Using the magnetic force, find the speed of the particle. From the centripetal force, findtheradius of the circle. Usingtheequation of period for circular motion, findtheperiod of motion.

Formulae are as follows:

$\text{F}=\text{qvB}$

role="math" localid="1663950112647" $\text{F}=\frac{{\text{mv}}^{\text{2}}}{\text{r}}$

Where F is force, B is the magnetic field, v is velocity, m is mass, q is the charge on particle, and r is the radius.

The particle is moving counterclockwise. From the right-hand rule, determine that the particle is an electron.

Speed of the particle:

Use the equation of magnetic force to find the speed of the particle.

$\text{F}=\text{qvB}$

Plugging the values,

$\begin{array}{c}3.20\times {10}^{-15}N=(1.602\times {10}^{-19}C)\times v\times (4.00\times {10}^{-3}T)\\ v=\frac{3.20\times {10}^{-15}N}{(1.602\times {10}^{-19}C)\times (4.00\times {10}^{-3}T)}\\ v=5.00\times {10}^{6}m/s\end{array}$

Hence, the speed of the particle is .$5.00\times {10}^{6}m/s$

The radius of the circle:

Centripetal force is provided by the magnetic force, so,

${\text{F}}_{\text{c}}=\text{qvB}$

$\frac{{\text{mv}}^{\text{2}}}{\text{r}}=\text{qvB}$

$\text{r}=\frac{\text{mv}}{\text{qB}}$

$\begin{array}{c}r=\frac{(9.11\times {10}^{-31}kg)(5.00\times {10}^{6}m/s)}{(1.602\times {10}^{-19}C)(4.00\times {10}^{-3}T)}\\ =7.11\times {10}^{-3}m\end{array}$

Hence, the radius of the circle is $7.11\times {10}^{-3}m$.

The period of the motion:

$\text{T}=\frac{\text{2πr}}{\text{v}}$

$\begin{array}{c}T=\frac{2\left(3.14\right)(7.11\times {10}^{-3}m)}{5.00\times {10}^{6}m/s}\\ =8.93\times {10}^{-9}s\end{array}$

Hence, the period of motion is$8.93\times {10}^{-9}s$.

Therefore, using the concept of magnetic force, centripetal force, and period, determine the speed, radius, and motion of the particle.