A proton is accelerated from rest through a potential difference of 150 V. It then enters a region of uniform magnetic field and moves in a circular path (radius = 12 cm). What is the magnitude of the magnetic field in this region?
A) 18 mT
B) 12 mT
C) 15 mT
D) 22 mT
E) 10 mT

Question

Quizplus · Accepted Answer

The kinetic energy gained by the proton when it is accelerated through a potential difference (V) is equal to the work done on it, which is given by $qV$, where $q$ is the charge of the proton (approximately $1.6 \times 10^{-19}$ C). This kinetic energy is then converted into centripetal force required for circular motion in the magnetic field, which is given by $\frac{mv^2}{r}$, where $m$ is the mass of the proton (approximately $1.67 \times 10^{-27}$ kg), $v$ is the velocity of the proton, and $r$ is the radius of the circular path. The magnetic force that provides this centripetal force is given by $qvB$, where $B$ is the magnetic field strength. Setting the centripetal force equal to the magnetic force gives $\frac{mv^2}{r} = qvB$, and solving for $B$ gives $B = \frac{mv}{qr}$.Using the kinetic energy equation $qV = \frac{1}{2}mv^2$ and solving for $v$, we can substitute $v$ into the equation for $B$. However, a simpler approach is to recognize that the kinetic energy gained ($qV$) is directly used to calculate $B$ without explicitly solving for $v$. Given:- $V = 150$ V- $r = 0.12$ m- $q = 1.6 \times 10^{-19}$ C- $m = 1.67 \times 10^{-27}$ kgThe correct calculation involves equating the kinetic energy to the work done by the magnetic field and solving for $B$, which yields a specific value for the magnetic field strength that ensures the proton moves in a circular path of the given radius. The correct answer, after performing the calculations with the given values, is found to be 15 mT, which corresponds to choice C. This involves using the relationship between the potential difference the proton is accelerated through, its charge, and the radius of the circular path in the magnetic field to find the magnetic field strength.

A Proton Is Accelerated from Rest Through a Potential Difference