Question

An electron is accelerated by a potential difference of 12000 volts. It then enters a uniform magnetic field of \[{10^{ - 3}}T\] applied perpendicular to the path of electrons. Find the radius of the path. Given mass of electron =\[9 \times {10^{ - 31}}\,Kg\] and charge on electron \[ = 1.6 \times {10^{ - 19}}\,C\]
A. 36.7 m
B. 36.7 cm
C. 3.67 m
D. 3.67 cm

Answer 1

Hint:When the electron is moving in a region with a magnetic field which is perpendicular to the motion of the electron. The magnetic force acts inward which makes the path of the motion circular.

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.
\[\dfrac{{m{v^2}}}{2} = qV\]
Here kinetic energy is equal to the work done by the potential difference.

Complete step by step solution:
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 applied potential difference is 12000 Volts.

Using the kinetic energy formula,
\[v = \sqrt {\dfrac{{2qV}}{m}} \]
The magnetic field strength is \[{10^{ - 3}}T\]
\[B = {10^{ - 3}}T\]
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
\[\dfrac{{m{v^2}}}{r} = evB \\ \]
\[\Rightarrow r = \dfrac{{m\sqrt {\dfrac{{2qV}}{m}} }}{{Bq}} \\ \]
\[\Rightarrow r = \dfrac{1}{B}\sqrt {\dfrac{{2Vm}}{q}} \]

Putting the values, we get the radius of the circular path is,
\[r = \dfrac{1}{{{{10}^{ - 3}}}}\sqrt {\dfrac{{2 \times 12000 \times 9 \times {{10}^{ - 31}}}}{{1.6 \times {{10}^{ - 19}}}}} m \\ \]
\[r = 36.7 \times {10^{ - 2}}m \\ \]
\[\therefore r = 36.7\,cm\]

Therefore, the correct option is B.

Note: In kinetic energy equivalence to potential energy formula, we take the magnitude of the charge. The work done by the potential difference applied offers the change in kinetic energy of the electron.