Question

Electrons move at right angles to a magnetic field of $1.5\times {{10}^{-2}}$Tesla with a speed of $6\times {{10}^{7}}m/s$. If the specific charge of the electron is $1.7\times {{10}^{11}}C/kg$. The radius of the circular path will be
A.$2.9cm$
B.$3.9cm$
C.$2.35cm$
D.$3cm$

Answer 1

Hint: The direction of the magnetic force on a freely charged particle in the uniform magnetic field is perpendicular to the directions of the velocity of the particle and this charged particle will follow a circular path within a uniform magnetic field. Relating the equation of centripetal force on a charged particle to the magnetic force equation, we can derive the formula of the radius of the circular path.


Formula used:
The radius of the circular path,$r=\dfrac{mv}{Bq}$
Here $m=$ mass of electrons
$v=$ velocity of electrons
$B=$magnetic field
$q=$charge of the particle


Complete answer:
When charged particles like electrons move through the uniform magnetic field, they experience a magnetic force. Consider a particle carrying charge $q$, that moves in the presence of a uniform magnetic field ($\vec{B}$) the magnetic force becomes perpendicular or right angles to the velocity $\vec{v}$ of the particle. This can be expressed mathematically:
$F=q(\vec{v}\times \vec{B})$
Or, $F=q(vB\sin \theta )$
Or,$F=q(vB\sin {{90}^{o}})$ [Since $\theta ={{90}^{o}}=$the angle between $\vec{B}\And \vec{v}$]
Or,$F=qvB$ …….(i)
 In this case, magnetic force becomes centripetal force due to its direction towards the circular path of the radius $r$of the charged particle.
$F=\dfrac{m{{v}^{2}}}{r}$ …….(ii)
By equating (i) & (ii) we get,
$qvB=\dfrac{m{{v}^{2}}}{r}$
Or,$r=\dfrac{mv}{qB}$ .…..(iii)
Given $\dfrac{q}{m}=1.7\times {{10}^{11}}C/kg$ , $B=1.5\times {{10}^{-2}}Tesla$, $v=6\times {{10}^{7}}m/s$
Putting the given values in equation (iii), we get
$r=\dfrac{6\times {{10}^{7}}m/s}{1.7\times {{10}^{11}}C/kg\times 1.5\times {{10}^{-2}}Tesla}=0.0235m$
Or,$r=0.0235\times 100=2.35cm$
Therefore, the radius of the circular path will be $2.35cm$.

Thus, option (C) is correct.




Note:To approach these types of questions we must take care about the formula and also whether the velocity of the particle is perpendicular or parallel to the magnetic field. This is because when the magnetic field and velocity are parallel to each other then the particle does not experience any magnetic force.