A proton, starting from rest, accelerates through a potential difference of $1.0 \mathrm { kV }$ and then moves into a magnetic field of $0.040 \mathrm {~T}$ at a right angle to the field. What is the radius of the proton's resulting orbit? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , m _ { \text {proton } } = 1.67 \times 10 ^ { - 27 } \mathrm {~kg} \right)$
A) $0.080 \mathrm {~m}$
B) $0.14 \mathrm {~m}$
C) $0.11 \mathrm {~m}$
D) $0.17 \mathrm {~m}$

Question

Quizplus · Accepted Answer

The radius $r$ of the proton's orbit in a magnetic field can be found using the formula $r = \frac{mv}{qB}$, where $m$ is the mass of the proton, $v$ is the velocity of the proton, $q$ is the charge of the proton, and $B$ is the magnetic field strength. The kinetic energy gained by the proton due to the acceleration through the potential difference $V$ is given by $K.E. = qV$, and since $K.E. = \frac{1}{2}mv^2$, we can solve for $v = \sqrt{\frac{2qV}{m}}$. Substituting the given values: $v = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \mathrm{C} \times 1000 \mathrm{V}}{1.67 \times 10^{-27} \mathrm{kg}}}$, and then substituting $v$ into the formula for $r$, we get $r = \frac{1.67 \times 10^{-27} \mathrm{kg} \times v}{1.60 \times 10^{-19} \mathrm{C} \times 0.040 \mathrm{T}}$. Calculating this gives a radius close to $0.11 \mathrm{m}$, making option C correct.

A Proton, Starting from Rest, Accelerates Through a Potential Difference 1.0kV1.0 \mathrm { kV }1.0kV

A Proton, Starting from Rest, Accelerates Through a Potential Difference $1.0 \mathrm { kV }$