Question

An \[\alpha \]- particle travels in a circular path of radius \[0.45{\rm{ }}m\] in a magnitude field \[B{\rm{ }} = {\rm{ }}1.2{\rm{ }}Wb/{m^2}\]with a speed of \[2.6{\rm{ }}\times\;{10^7}\;m/s\]. Find the period of revolution of the \[\alpha \] - particle.
A. \[1.1 \times {10^{ - 5}}s\;\]
B. \[1.1 \times {10^{ - 6}}s\;\]
C. \[1.1 \times {10^{ - 7}}s\;\]
D. \[1.1 \times {10^8}s\;\]

Answer 1

Hint: In the given question, we need to find the period of revolution of the \[\alpha \] - particle. For this, we need to use the formula for force experienced by a charged particle in an external magnetic field to get the desired result.


Formula used:
The following formula is used for solving the given question.
The period of revolution of the \[\alpha \] - particle is given by
 \[T = \dfrac{{2\pi r}}{v}\]
Here, \[T\] is the period, \[r\] is the radius, and \[v\] is the velocity.






Complete answer:
We know that the period of revolution of the \[\alpha \] - particle is given by
\[T = \dfrac{{2\pi r}}{v}\]
Here, \[T\]is the period, \[r\] is the radius, and \[v\]is the velocity.
But \[r = 0.45{\rm{ }}m\],and \[v = 2.6{\rm{ }}\times\;{10^7}\;m/s\]
So, we get
 \[T = \dfrac{{2\pi \times 0.45}}{{2.6{\rm{ }}\times\;{{10}^7}}}\]
By simplifying, we get
\[T = \dfrac{{2\pi \times 0.45}}{{2.6{\rm{ }}\times\;{{10}^7}}}\]
This gives \[1.1 \times {10^{ - 7}}s\;\]
Hence, the period of revolution of the \[\alpha \] - particle is \[1.1 \times {10^{ - 7}}s\;\].

Therefore, the correct option is (C).




Note: If a proton and alpha particle enters a uniform magnetic field perpendicularly at the exact same speed, the alpha particle's time period of revolution is twice that of the proton. Many students make mistakes in the calculation part as well as simplifying the power of ten. This is the only way through which we can solve the example in the simplest way. Also, it is essential to do calculations carefully to get the correct value of the period of revolution of the \[\alpha \] - particle.