Question

An electron beam travels with a velocity ${{1}}{{.6 \times 1}}{{{0}}^{{7}}}\dfrac{{{m}}}{{{{{s}}^{{{ - 3}}}}}}$, perpendicularly to magnetic field of intensity ${{0}}{{.1T}}{{.}}$ The radius of the path of electron beam $\left( {{{{m}}_{{e}}}{{ = 9 \times 1}}{{{0}}^{{{ - 31}}}}{{kg}}} \right)$
A) ${{9 \times 1}}{{{0}}^{{{ - 5}}}}{{m}}$
B) ${{9 \times 1}}{{{0}}^{{{ - 4}}}}{{m}}$
C) ${{9 \times 1}}{{{0}}^{{{ - 3}}}}{{m}}$
D) ${{9 \times 1}}{{{0}}^{{{ + 3}}}}{{m}}$

Answer 1

Hint: The question is based on the concept of movements of charged particles in a given uniform magnetic field. If the velocity vector of the charged particle is perpendicular to the applied magnetic field the charged particle performs circular motion.

Formula used:
${{R = }}\dfrac{{{{m \times V}}}}{{{{QB}}}}$ where, ${{R = }}$Radius of the circular path followed by the charged particle.
${{m = }}$ mass of the charged particle
${{V = }}$ Velocity of the charged particle.
${{Q = }}$ charge on the charged particle, ${{B = }}$magnetic field

Complete step by step solution:
We know that when a charged particle enters a magnetic field region with its velocity vector perpendicular to the magnetic field it will perform the circular motion. The necessary centripetal force for this motion is given by the magnetic force which is applied on the charged particle due to the magnetic field.
The expression for the magnetic force${{ = }}\mathop {{F}}\limits^ \to {{ = Q}}\left( {\mathop {{V}}\limits^ \to } \right)\left( {\mathop {{B}}\limits^ \to } \right)$
\[\mathop {{F}}\limits^ \to {{ = Q}}\left( {\mathop {{V}}\limits^ \to {{ \times }}\mathop {{B}}\limits^ \to } \right)\]
As velocity vector is perpendicular to the magnetic field vector, the magnetic of the force becomes
\[{{F = QVB}}\]
This force supplies the necessary centripetal force for the circular motion of the changed particle.
So ${{QVB = }}\dfrac{{{{m}}{{{V}}^{{2}}}}}{{{R}}}$
Solving this ${{R = }}\dfrac{{{{mV}}}}{{{{QB}}}}$
Given in the question,
Mass of electron, ${{m}}$= ${{9 \times 1}}{{{0}}^{{{ - 31}}}}{{kg}}$
Charge of electron, ${{Q = 1}}{{.6 \times 1}}{{{0}}^{{{ - 19}}}}{{C}}$
Velocity of electron, ${{V = 1}}{{.6 \times 1}}{{{0}}^{{7}}}\dfrac{{{m}}}{{{{{s}}^{{{ - 3}}}}}}$
Intensity of magnetic field, ${{B = 0}}{{.1T}}$
Putting the values in the formula
${{R = }}\dfrac{{{{9 \times 1}}{{{0}}^{{{ - 31}}}}{{ \times 1}}{{.6 \times 1}}{{{0}}^{{7}}}}}{{{{1}}{{.6 \times 1}}{{{0}}^{{{ - 19}}}}{{ \times 0}}{{.1}}}} = 0.9$ $mm$

Hence, Option (B) is correct.

Additional Information:
It should be remembered that when a charged particle enters a magnetic field region it will perform circular motion only when the velocity vector is perpendicular to the magnetic field vector.
If supposing that the magnetic field vector is no longer perpendicular to the velocity vector, then the path followed by the charged particle will no longer be circular, but it will be helical in shape so care should be taken to check whether the velocity vector is perpendicular to the magnetic field vector or not.

Note: Whenever the Velocity vector is perpendicular to the magnetic field vector with their magnitudes uniform then the charged particle will perform uniform circular motion.