Question

j j thomson is best known for his discoveries about the nature of cathode rays. his other important contribution was the invention, together with one of his students, of the
part a
after being accelerated to a speed of 1.92×10^5 m/s, the particle enters a uniform magnetic field of strength 0.900 t and travels in a circle of radius 31.0 cm (determined by observing where it hits the screen as shown in the figure). the results of this experiment allow one to find m/q
find the ratio m/q for this particle.
express your answer numerically in kilograms per coulomb.
view available hint(s)
m/q = kg/c

Explanation:

Step1: Identify the centripetal - force formula

The magnetic force $F = qvB$ provides the centripetal force $F_c=\frac{mv^{2}}{R}$ for the charged - particle moving in a magnetic field. So, $qvB=\frac{mv^{2}}{R}$.

Step2: Solve for $\frac{m}{q}$

From $qvB=\frac{mv^{2}}{R}$, we can cross - multiply to get $qBR = mv$. Then, $\frac{m}{q}=\frac{BR}{v}$.

Step3: Substitute the given values

Given $B = 0.900\ T$, $R=31.0\ cm = 0.310\ m$, and $v = 1.92\times10^{5}\ m/s$. Substitute these values into the formula $\frac{m}{q}=\frac{BR}{v}$.
$\frac{m}{q}=\frac{0.900\times0.310}{1.92\times10^{5}}$
$\frac{m}{q}=\frac{0.279}{1.92\times10^{5}}$
$\frac{m}{q}=1.45\times10^{-6}\ kg/C$

Answer:

$1.45\times10^{-6}$