Answer
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Hint: When $\theta = {90^ \circ }$ i.e., the direction of velocity of charge is perpendicular to direction of magnetic field and the force on moving charge will be maximum. We can calculate this by using the Lorentz equation of force
Complete step by step answer:
The Lorentz force is the cause of the electromagnetic field. It can also be said as an electromagnetic force. We can say that the Lorentz force can be defined as the combination of electric and magnetic force on a point charge due to electromagnetic fields.
When a object is travelling in an electric field $E$ having charge $q$ and is moving with velocity $v$, then, the force experienced by that object in magnetic field $B$ will be –
$
F = qE + q\left( {v \times B} \right) \\
F = q\left( {E + v \times B} \right) \\
$
For the continuous charge distribution, the Lorentz force can be illustrated as –
$dF = dq\left( {E + v \times B} \right)$
Where, $dF$ is the force on a small piece of charge
$dq$is the charge of a small piece.
When the object is not travelling with electric field but have charge $q$ and moves with velocity $v$, then the force experienced by that object in magnetic field $B$ will be –
$F = qvB\sin \theta $
This is the Lorentz force due to magnetic force.
When direction of velocity of charge is parallel to magnetic field then, $\theta = {0^ \circ }$ or ${180^ \circ }$
$F = qvB\sin {0^ \circ }$
$\because \sin {0^ \circ } = 0$
So, force will also be zero. Hence, this will be the minimum value
When direction of velocity of charge is perpendicular to magnetic field, $\theta = {90^ \circ }$.
$\therefore F = qvB\sin {90^ \circ }$
$\because \sin {90^ \circ } = 1$
$\therefore F = qvB$
This is the maximum value of force when the object having charge $q$ moving with velocity $v$ experiences magnetic field $B$.
And also, it is proved that force is not independent of magnetic field $B$ in any case.
Hence, both Assertion and Reason are not correct.
So, the correct option is (D).
Note: The implications of the Lorentz force are:
1) The force in this relationship is the vector product as it has both magnitude and direction.
2) In this relationship, the direction of force is given by the right hand thumb rule.
3)The force is perpendicular to the velocity $v$ of the charge $q$ and the magnetic field $B$.
Complete step by step answer:
The Lorentz force is the cause of the electromagnetic field. It can also be said as an electromagnetic force. We can say that the Lorentz force can be defined as the combination of electric and magnetic force on a point charge due to electromagnetic fields.
When a object is travelling in an electric field $E$ having charge $q$ and is moving with velocity $v$, then, the force experienced by that object in magnetic field $B$ will be –
$
F = qE + q\left( {v \times B} \right) \\
F = q\left( {E + v \times B} \right) \\
$
For the continuous charge distribution, the Lorentz force can be illustrated as –
$dF = dq\left( {E + v \times B} \right)$
Where, $dF$ is the force on a small piece of charge
$dq$is the charge of a small piece.
When the object is not travelling with electric field but have charge $q$ and moves with velocity $v$, then the force experienced by that object in magnetic field $B$ will be –
$F = qvB\sin \theta $
This is the Lorentz force due to magnetic force.
When direction of velocity of charge is parallel to magnetic field then, $\theta = {0^ \circ }$ or ${180^ \circ }$
$F = qvB\sin {0^ \circ }$
$\because \sin {0^ \circ } = 0$
So, force will also be zero. Hence, this will be the minimum value
When direction of velocity of charge is perpendicular to magnetic field, $\theta = {90^ \circ }$.
$\therefore F = qvB\sin {90^ \circ }$
$\because \sin {90^ \circ } = 1$
$\therefore F = qvB$
This is the maximum value of force when the object having charge $q$ moving with velocity $v$ experiences magnetic field $B$.
And also, it is proved that force is not independent of magnetic field $B$ in any case.
Hence, both Assertion and Reason are not correct.
So, the correct option is (D).
Note: The implications of the Lorentz force are:
1) The force in this relationship is the vector product as it has both magnitude and direction.
2) In this relationship, the direction of force is given by the right hand thumb rule.
3)The force is perpendicular to the velocity $v$ of the charge $q$ and the magnetic field $B$.
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