Question

An electron has mass\[9 \times {10^{ - 31}}kg\] and charge\[1.6 \times {10^{ - 19}}C\] is moving with velocity of\[{10^6}m/s\] enters a region where magnetic field exists. If it describes a circle of radius 0.10 m, the intensity of magnetic field must be
A.\[1.8 \times {10^{ - 4}}T\]
B.\[5.6 \times {10^{ - 5}}T\]
C.\[14.4 \times {10^{ - 5}}T\]
D.\[1.3 \times {10^{ - 6}}T\]

Answer 1

Hint: Given that the electron is moving with some velocity, it will be affected by the magnetic field. This is because the magnetic field exerts a force on the charge when it is moving. The force experienced by a moving charge depends on many factors.




Formula used:
The intensity of the magnetic field can be known by using the following formula written,
\[R = \dfrac{{mv}}{{qB}}\]



Complete answer:
Given that the mass of an electron is\[m = 9 \times {10^{ - 31}}kg\]
Given that the charge on the electron is\[q = 1.6 \times {10^{ - 19}}C\]
Also given that the velocity of moving electron is\[v = {10^6}m/s\]
Given that the radius of the circle is, R=0.10m
The intensity of the magnetic field can be known by using the following formula written,
\[R = \dfrac{{mv}}{{qB}}\]
Substituting all the values in the above written formula and solving for the value of intensity of magnetic field,
\[B = \dfrac{{9 \times {{10}^{ - 31}} \times {{10}^6}}}{{1.6 \times {{10}^{ - 19}} \times 0.1}}\]
On solving the above equation, the value of magnetic field will be
\[B = 5.6 \times {10^{ - 5}}T\]
Therefore, the intensity of the magnetic field will be\[5.6 \times {10^{ - 5}}T\]

Hence, Option B is the correct answer.






Note: If the particle is moving in the direction of the magnetic field with some velocity, then the velocity of the charged particle will be perpendicular to the magnetic field. As a result, the particle will start moving in a circular motion as the magnetic force becomes centripetal. On the other hand, if the velocity and the magnetic field are in the same direction then the charged particle will not experience any force and the magnitude of the charged particle will remain the same. As a result of both the components, the force acts on the charged particle and it starts moving in a helical path.