Question

An electron having charge \[1.6 \times {10^{ - 19}}C\]and mass \[9 \times {10^{ - 31}}Kg\]is moving with \[4 \times {10^6}m{s^{ - 1}}\]speed in a magnetic field \[2 \times {10^{ - 1}}T\]in a circular orbit. The force acting on electron and the radius of the circular orbit will be
A. \[12.8 \times {10^{ - 13}}N,1.1 \times {10^{ - 4}}m\]
B. \[1.28 \times {10^{ - 14}}N,1.1 \times {10^{ - 3}}m\]
C. \[1.28 \times {10^{ - 13}}N,1.1 \times {10^{ - 3}}m\]
D. \[1.28 \times {10^{ - 13}}N,1.1 \times {10^{ - 4}}m\]

Answer 1

Hint: When the electron is moving in a region with a magnetic field which is perpendicular to the motion of the electron then the path of the motion is circular. The magnetic force is perpendicular to the magnetic field and the velocity. At equilibrium along the radius of the circular path, the outward centrifugal force must be balanced by the magnetic force which is inward.


Formula used:
\[{F_m} = qVB\], here \[{F_m}\]is the magnetic field acting on the charge particle q moving with speed v in a magnetic field strength B where the motion is perpendicular to the magnetic field.
\[{F_c} = \dfrac{{m{v^2}}}{r}\], here \[{F_c}\]is the centrifugal force acting on a body of mass m moving with speed v in a circular path of radius r.




Complete answer:
The mass of the electron is given as \[9 \times {10^{ - 31}}Kg\]
\[m = 9 \times {10^{ - 31}}Kg\]
The charge on the electron is \[1.6 \times {10^{ - 19}}C\]
\[q = 1.6 \times {10^{ - 19}}C\]
The speed of the electron is \[4 \times {10^6}m{s^{ - 1}}\]
\[v = 4 \times {10^6}m{s^{ - 1}}\]
The magnetic field strength is \[2 \times {10^{ - 1}}T\]
\[B = 2 \times {10^{ - 1}}T\]
The magnetic force acting on the moving charge is,
\[{F_m} = qvB\]
\[{F_m} = 1.6 \times {10^{ - 19}} \times 4 \times {10^6} \times 2 \times {10^{ - 1}}N\]
\[{F_m} = 1.28 \times {10^{ - 13}}N\]
Then the centrifugal force acting on the electron is,
\[{F_C} = \dfrac{{m{v^2}}}{r} \ldots \ldots \left( i \right)\]
To balance the outward centrifugal force, the magnetic force on the electron should be inward and have magnitude equal to the outward force.
The magnetic force acting on the moving charge is,
\[{F_m} = qvB\]
Putting the charge of the electron in the formula, we get
\[{F_m} = qvB \ldots \ldots \left( {ii} \right)\]
For the electron to be in circular motion, the two forces must balance each other.
On balancing the force, we get
\[\dfrac{{m{v^2}}}{r} = evB\]
\[r = \dfrac{{mv}}{{Bq}}\]
Putting the values, we get the radius of the circular path is,
\[r = \dfrac{{9 \times {{10}^{ - 31}} \times 4 \times {{10}^6}}}{{2 \times {{10}^{ - 1}} \times 1.6 \times {{10}^{ - 19}}}}m\]
\[r = 1.1 \times {10^{ - 4}}m\]

Therefore, the correct option is (D).



Note:In the circular path, the perpendicular component of the magnetic field to the velocity is the cause of the centripetal force which determines the radius of the circular path.