
Write the expression for Lorentz magnetic force on a particle of charge $q$ moving with velocity $v$ in a magnetic field $B$. Shown that two no work is done by this force on the charged particle.
Answer
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Hint
Lorentz force, the force exerted on a charged particle $q$ moving with velocity $v$ through an electric field $E$ and magnetic field $B$. The entire electromagnetic force $F$ on the charged particle is called the Lorentz force.
Complete step by step answer
We know that,
Lorentz force = magnetic force + electric force.
So, now we can say,
$F{\text{ }} = {\text{ }}[{\text{ }}qvb{\text{ }}sin\theta \; + {\text{ }}qe{\text{ }}]$
$ \Rightarrow \vec F = q(\vec V \times \vec B)\;d\vec s$
Now, $\vec F$ is perpendicular to both $\vec V$ and $\vec B$.
If $d\vec s$ is the instantaneous displacement of the change-
Then, $d\vec s$ is also perpendicular to $\vec F$
Now, according to work done formula,
$W = \vec F.d\vec s$
$ \Rightarrow W = Fs\cos {90^0 }$
But, the value of $cos 90^0$ is equal to zero.
So, $W = 0$,
That means the work done is zero and the increase in kinetic energy is zero.
Note
The work is done when a force acts upon an object to cause a displacement. Three quantities must be known in order to calculate the amount of work. Those three quantities are force, displacement and the angle between the force and the displacement.
Lorentz force, the force exerted on a charged particle $q$ moving with velocity $v$ through an electric field $E$ and magnetic field $B$. The entire electromagnetic force $F$ on the charged particle is called the Lorentz force.
Complete step by step answer
We know that,
Lorentz force = magnetic force + electric force.
So, now we can say,
$F{\text{ }} = {\text{ }}[{\text{ }}qvb{\text{ }}sin\theta \; + {\text{ }}qe{\text{ }}]$
$ \Rightarrow \vec F = q(\vec V \times \vec B)\;d\vec s$
Now, $\vec F$ is perpendicular to both $\vec V$ and $\vec B$.
If $d\vec s$ is the instantaneous displacement of the change-
Then, $d\vec s$ is also perpendicular to $\vec F$
Now, according to work done formula,
$W = \vec F.d\vec s$
$ \Rightarrow W = Fs\cos {90^0 }$
But, the value of $cos 90^0$ is equal to zero.
So, $W = 0$,
That means the work done is zero and the increase in kinetic energy is zero.
Note
The work is done when a force acts upon an object to cause a displacement. Three quantities must be known in order to calculate the amount of work. Those three quantities are force, displacement and the angle between the force and the displacement.
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