The dimensional formula for the magnetic fields is:
A) $[M{T^{ - 2}}{A^{ - 1}}]$
B) $[M{L^2}{T^{ - 1}}{A^{ - 2}}]$
C) $[M{T^{ - 2}}{A^{ - 2}}]$
D) \[[M{T^{ - 1}}{A^{ - 2}}]\]
Answer
257.1k+ views
Hint: In order to solve this question you have to know the formula for finding the magnetic field. Further, you have to know the dimension for all the terms present in that formula that are force, velocity and charge. Also remember that the $\sin \theta $ is dimensionless.
Formula used:
The Lorentz force is given by,
$F = qvB\sin \theta $
Where, $q$ is the electric charge of the particle
$v$ is the velocity
$B$ is the magnetic field
Complete step by step solution:
We know that the Lorentz force is given by,
$F = qvB\sin \theta $…….(i)
As we also know that force is given by,
$F = ma$ , where $m$ is the mass of the body and $a$ is the acceleration
So, the dimensional formula for force is given by
Dimension of force, $F = [ML{T^{ - 2}}]$
Since, the velocity is given by
$v = \dfrac{d}{t}$ , where $d$ is the distance and $t$ is the time
So, the dimensional formula for velocity is given by,
Dimension of velocity, $v = [L{T^{ - 1}}]$
Since, the electric charge is given by,
$q = I \times t$
So, the dimensional formula for the electric charge is given by,
Dimension of electric charge, $q = [AT]$
And we know that the term $\sin \theta $ is dimensionless.
Thus from equation (i), the formula for magnetic field is given by
Magnetic field, $B = \dfrac{F}{{qv\sin \theta }}$
Now put the dimensional formula for each terms in the above equation, we get
$ \Rightarrow B = \dfrac{{[ML{T^{ - 2}}]}}{{[L{T^{ - 1}}][AT]}}$ {since $\sin \theta $ is dimensionless}
On further solving, we get
$ \Rightarrow B = [M{T^{ - 2}}{A^{ - 1}}]$
Hence the dimensional formula for the magnetic fields is $[M{T^{ - 2}}{A^{ - 1}}]$.
Therefore, the correct option is (A).
Note: Lorentz force is the combination of two forces on a point charge due to electromagnetic fields, that forces are the magnetic and electric force. It is used in electromagnetism and is also known as the electromagnetic force. A magnetic field is like an invisible space around a magnetic object. Basically, a magnetic field is used to describe the distribution of magnetic force around a magnetic object.
Formula used:
The Lorentz force is given by,
$F = qvB\sin \theta $
Where, $q$ is the electric charge of the particle
$v$ is the velocity
$B$ is the magnetic field
Complete step by step solution:
We know that the Lorentz force is given by,
$F = qvB\sin \theta $…….(i)
As we also know that force is given by,
$F = ma$ , where $m$ is the mass of the body and $a$ is the acceleration
So, the dimensional formula for force is given by
Dimension of force, $F = [ML{T^{ - 2}}]$
Since, the velocity is given by
$v = \dfrac{d}{t}$ , where $d$ is the distance and $t$ is the time
So, the dimensional formula for velocity is given by,
Dimension of velocity, $v = [L{T^{ - 1}}]$
Since, the electric charge is given by,
$q = I \times t$
So, the dimensional formula for the electric charge is given by,
Dimension of electric charge, $q = [AT]$
And we know that the term $\sin \theta $ is dimensionless.
Thus from equation (i), the formula for magnetic field is given by
Magnetic field, $B = \dfrac{F}{{qv\sin \theta }}$
Now put the dimensional formula for each terms in the above equation, we get
$ \Rightarrow B = \dfrac{{[ML{T^{ - 2}}]}}{{[L{T^{ - 1}}][AT]}}$ {since $\sin \theta $ is dimensionless}
On further solving, we get
$ \Rightarrow B = [M{T^{ - 2}}{A^{ - 1}}]$
Hence the dimensional formula for the magnetic fields is $[M{T^{ - 2}}{A^{ - 1}}]$.
Therefore, the correct option is (A).
Note: Lorentz force is the combination of two forces on a point charge due to electromagnetic fields, that forces are the magnetic and electric force. It is used in electromagnetism and is also known as the electromagnetic force. A magnetic field is like an invisible space around a magnetic object. Basically, a magnetic field is used to describe the distribution of magnetic force around a magnetic object.
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