Question

The $Bqv$ has dimensions
A. $[{M^1}{L^1}{T^{ - 2}}]$
B. $[{M^2}{L^2}{T^{ - 2}}]$
C. $[{M^1}{L^1}{T^{ - 3}}]$
D. Cannot be expressed in terms of $[MLT]$

Answer 1

Hint: The term given above is the formula known as Lorentz force law which is given as $F = qvB\sin \theta $. But in the question, the formula of the force is given as $F = qvB$. This means that the angle between the motion of charge and the magnetic field is $90^\circ $, that is why the force will become $F = qvB$.

Formula used:
The Lorentz formula law given in the question is given below
$F = qvB\sin \theta $
Here, $F$ is the force acting on the charge, $q$ is the charge, $v$ is the velocity of the charge, $B$ is the magnetic field acting on the charge and $\theta $ is the angle between the velocity of the charge and the magnetic field.

Complete step by step answer:
The formula of force given in the question is given below
$F = qvB$
Here, $F$ is the force acting on the charge, $q$ is the charge, $v$ is the velocity of the charge and $B$ is the magnetic field acting on the charge.
Now, we know that the unit of charge is $Coulomb$ whose dimensional formula is $[{M^0}{L^0}{T^1}{I^1}]$. Also, the unit of velocity will be $m{s^{ - 1}}$, therefore, the dimensional formula will be $[{M^0}L{T^{ - 1}}]$. The unit of magnetic field is $Tesla$ which can also be written as $\dfrac{N}{{C\,m{s^{ - 1}}}}$, therefore, the dimensional will be $[{M^1}{L^0}{T^{ - 2}}{I^{ - 1}}]$.
Putting these values in the formula of force, we get
$qvB = \left[ {{M^0}{L^0}{T^1}{I^1}} \right]\left[ {{M^0}L{T^{ - 1}}} \right]\left[ {{M^1}{L^0}{T^{ - 2}}{I^{ - 1}}} \right]$
$ \therefore \,qvB = [{M^1}{L^1}{T^{ - 2}}]$
Therefore, $Bqv$ has dimensions $[{M^1}{L^1}{T^{ - 2}}]$.

Hence, option A is the correct.

Note:An alternate method to solve the question is given below.
The formula of Lorentz force law when the angle between the motion of charge and the magnetic field is $90^\circ $ is given below
$F = qvB$
Also, the formula of Newton’s second law of motion is given below
$F = ma$
Now, equating both the equations, we get
$qvB = ma$
Now, we know that the unit of acceleration is $m{s^{ - 2}}$, therefore, the dimensional formula will be $L{T^{ - 2}}$. Also, the dimensional formula of mass is $M$.
Therefore, the above equation will become
$ \Rightarrow \,qvB = \left[ M \right]\left[ {L{T^{ - 2}}} \right]$
$ \therefore \,qvB = \left[ {ML{T^{ - 2}}} \right]$
Which is the required answer.