Question

In Figure 28-39, a charged particle moves into a region of uniform magnetic field, goes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It spends 130 ns in the region. (a)What is the magnitude of $\overrightarrow{\mathbf{B}}$?

(b)If the particle is sent back through the magnetic field (along the same initial path) but with 2.00 times its previous kinetic energy, how much time does it spend in the field during this trip?

Short answer

a) The magnitude of the magnetic field is$0.252T$ .

b) Time spent is $130ns$.

Step-by-step solution

Listing the given quantities

Time spent by the particle is $130ns=1.30\times {10}^{-7}s$

Understanding the concept of centrifugal force

We have to use the formula for magnetic force and centrifugal force to find the magnetic field.Also, we need to express the period in terms of the magnetic field to determine the time spent by a particle in the region.

Formula:

$F_m=qVB$

$F_c=\frac{m{v}^2}{r}$

Step 3:(a) Calculation of the magnetic field

The particle is pointing downward while entering the magnetic field, and the direction of the magnetic field is out of the page, as shown in the figure. So, according to the right-hand rule,$\overrightarrow{v}\times \overrightarrow{B}$

points towards the left. As this is the direction of a charged particle, we can conclude that the particle has a positive charge. Hence, it is a proton.

Now, we have:

$\begin{array}{c}qvB=\frac{m{v}^2}{r}\\ qB=m\omega \\ \omega =\frac{qB}{m}\\ \frac{2\pi }{T}=\frac{qB}{m}\end{array}$

So,

$B=\frac{2\pi m}{qT}$

Here, T is the period given by:

,

$T=2\times t=2\times 1.30\times {10}^{-7}\text{\hspace{0.17em}s}$

So, Magnetic force can be calculated as:

$\begin{array}{c}B=\frac{2\pi \times 1.67\times {10}^{-27}}{2\times 1.6\times {10}^{-19}\times 1.30\times {10}^{-7}}.\\ =0.252T\end{array}$

Thus, the magnitude of the magnetic field is $0.252T$.

Step 4:(b) Calculation of the time spent

We have the equation for the time period as follows:

$T=\frac{2\pi m}{qB}$

This shows that period does not depend on speed, so it will remain thesame even though kinetic energy is double.

Therefore, time spent in the magnetic field region will remain .$130ns$$130ns$