Question

One proton beam enters a magnetic field of \[{10^{ - 4}}T\]normally, Specific charge \[ = {10^{11}}C/kg\],velocity \[{10^7}m/s\]. What is the radius of the circle described by it
A. 0.1m
B. 1m
C. 10m
D. None of these

Answer 1

Hint: When the proton enters into a magnetic field then it experiences the magnetic force due to interaction between the charge and the magnetic field. The magnetic force is acting perpendicular to the direction of the motion of the proton and hence the path is circular in motion.


Formula used:
\[r = \dfrac{{mv}}{{Bq}}\], here r is the radius of the circular path when a charged particle of mass m and charge q enters into a region of magnetic field strength B with speed v.




Complete answer:
The specific charge of the proton is the ratio of charge to the mass of the proton,
\[\dfrac{q}{m} = {10^{11}}C/kg\]
The magnetic field strength is given as \[{10^{ - 4}}T\]
\[B = {10^{ - 4}}T\]
The speed of the proton is given as \[{10^7}m/s\]
\[v = {10^7}m/s\]
The magnitude of the centripetal force is equal to the magnetic force magnitude,
\[{F_r} = qvB\]
The outward centrifugal force acting on the proton in the circular path is,
\[{F_C} = \dfrac{{m{v^2}}}{r}\]
At radial equilibrium state, the outward force is balanced by the inward force;
\[\dfrac{{m{v^2}}}{r} = qvB\]
\[r = \dfrac{{mv}}{{Bq}}\]
\[r = \dfrac{v}{{B\left( {\dfrac{q}{m}} \right)}}\]
Putting the values, we get
\[r = \dfrac{{{{10}^7}}}{{{{10}^{ - 4}} \times {{10}^{11}}}}m\]
\[r = 1.0m\]
Hence, the radius of the circular path of the proton is 1.0m.
Therefore, the correct option is (B).





Note:We should be careful about the charge to mass ratio as a specific charge and use it in the formula obtained for the radius of the circular path of the charge when moving in the region of magnetic field.