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? hint: do you recall that when the velocity vector of a charged particle is perpendicular to a uniform magnetic field, a magnetic force makes the particle move in circular motion at constant speed, experiencing a centripetal acceleration? how do you relate the force and the acceleration to get the magnetic field? etextbook and media assistance used (a) number units t (b) number 146 units ns

Explanation:

Step1: Recall centripetal - force formula for charged particle in magnetic field

The centripetal force for a charged particle moving in a magnetic field is given by $F = qvB$, and also $F=ma_c=\frac{mv^{2}}{r}$, so $qvB=\frac{mv^{2}}{r}$, which simplifies to $r = \frac{mv}{qB}$. The time spent in the magnetic - field for half - circle is $t=\frac{\pi r}{v}$. Combining these equations, we get $t=\frac{\pi m}{qB}$.

Step2: Determine the mass and charge for proton and electron

The mass of a proton is $m_p = 1.67\times10^{- 27}\text{ kg}$ and its charge is $q_p=1.6\times10^{-19}\text{ C}$, the mass of an electron is $m_e = 9.11\times10^{-31}\text{ kg}$ and its charge is $q_e=- 1.6\times10^{-19}\text{ C}$. Since we are dealing with magnitudes of forces and time, we can ignore the sign of the electron's charge.

Step3: Calculate the magnetic field magnitude (a)

We know $t = 146\text{ ns}=146\times10^{-9}\text{ s}$. For a charged particle moving in a semi - circular path in a magnetic field $t=\frac{\pi m}{qB}$. Rearranging for $B$, we have $B=\frac{\pi m}{qt}$. Let's assume it's a proton (since proton is much heavier than electron and more likely to have the given time value considering the mass - to - charge ratio). Substituting $m = m_p = 1.67\times10^{-27}\text{ kg}$, $q = 1.6\times10^{-19}\text{ C}$ and $t = 146\times10^{-9}\text{ s}$ into the formula:
\[B=\frac{\pi\times1.67\times10^{-27}}{1.6\times10^{-19}\times146\times10^{-9}}\]
\[B=\frac{3.14\times1.67\times10^{-27}}{1.6\times146\times10^{-28}}\]
\[B=\frac{5.2438\times10^{-27}}{233.6\times10^{-28}}\]
\[B = 0.225\text{ T}\]

Step4: Calculate the time for the second case (b)

If the kinetic energy $K' = 3.22K$. Since $K=\frac{1}{2}mv^{2}$, if $K' = 3.22K$, then $\frac{1}{2}mv'^{2}=3.22\times\frac{1}{2}mv^{2}$, so $v'=\sqrt{3.22}v$. The radius of the circular path $r=\frac{mv}{qB}$, and the time for a semi - circular path $t=\frac{\pi r}{v}=\frac{\pi m}{qB}$ (independent of velocity). So the time spent in the field remains the same, $t = 146\text{ ns}$.

Answer:

(a) $0.225\text{ T}$
(b) $146\text{ ns}$