Question

in the figure, a charged particle moves into a region of uniform magnetic field b, 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 146 ns in the region. (a) what is the magnitude of b? (b) if the particle is sent back through the magnetic field (along the same initial path) but with 3.22 times its previous kinetic energy, how much time does it spend in the field during this trip? (a) number units (b) number units

Explanation:

Step1: Recall the formula for time - of - flight in magnetic field for a semi - circular path

The time taken by a charged particle to move through a semi - circular path in a magnetic field is given by $t=\frac{\pi m}{qB}$, and the kinetic energy is $K = \frac{1}{2}mv^{2}$. The centripetal force is provided by the magnetic force $qvB=\frac{mv^{2}}{r}$, so $r=\frac{mv}{qB}$. The time taken to traverse a semi - circle $t=\frac{\pi r}{v}=\frac{\pi m}{qB}$.

Step2: Analyze the relationship between kinetic energy and time

We know that $K=\frac{1}{2}mv^{2}$ and $r = \frac{mv}{qB}$. If the new kinetic energy $K_{new}=3.22K_{old}$, then $\frac{1}{2}mv_{new}^{2}=3.22\times\frac{1}{2}mv_{old}^{2}$, so $v_{new}=\sqrt{3.22}v_{old}$. Also, $r=\frac{mv}{qB}$, and $t=\frac{\pi r}{v}=\frac{\pi m}{qB}$ which is independent of the speed of the particle.

(a)
The time taken by a charged particle to move through a semi - circular path in a magnetic field $t=\frac{\pi m}{qB}$ is independent of the speed of the particle. Given $t = 146$ ns.
The magnitude of the magnetic field $B$ can be found from $t=\frac{\pi m}{qB}$. For a proton $m = 1.67\times10^{- 27}$ kg and $q=1.6\times10^{-19}$ C.
We know $t=\frac{\pi m}{qB}$, so $B=\frac{\pi m}{qt}$.
Substituting the values:
\[

$$\begin{align*} B&=\frac{\pi\times1.67\times 10^{-27}}{1.6\times10^{-19}\times146\times10^{-9}}\\ &=\frac{\pi\times1.67\times 10^{-27}}{2.336\times10^{-17}}\\ &\approx2.23\times 10^{-10}\text{ T} \end{align*}$$

\]

(b)
Since the time taken to traverse a semi - circular path in a magnetic field $t=\frac{\pi m}{qB}$ is independent of the kinetic energy of the particle (because $t$ does not depend on $v$ as $t=\frac{\pi r}{v}$ and $r = \frac{mv}{qB}$), the time spent in the field during the second trip is the same as the first trip when the path is the same. So $t = 146$ ns.

Answer:

(a) Number: $2.23\times 10^{-10}$
Units: T
(b) Number: $146$
Units: ns