
A stationary magnet does not interact with:
A. iron rod
B. Moving charge
C. Moving magnet
D. Stationary charge
Answer
511.8k+ views
Hint: As we all know that the whole concept of magnetism is arbitrary, when the charge moves with some speed relative to the speed of light in a vacuum then the separation between the charged particles is greatly reduced which turns the conductor more charged and which thereby attracts other charged particles.
Complete step by step solution:
We can recall that the magnetic force on a charged particle is orthogonal to the magnetic field and we can represent it by the formula as,
$ \Rightarrow F = qvB\sin \theta $
In the above formula, $B$ is the magnetic field vector, v is the velocity of the particle, and $\theta $ is the angle between the magnetic field and particle velocity. The Force due to the magnetic field on a moving charged particle is perpendicular to both the velocity plane and magnetic field plane.
Therefore, we can say that if the motion of the charged particle is parallel to the magnetic field then the force due to the magnetic field on a charged particle is zero. This can be illustrated by the above formula as,
\[ \Rightarrow F = qvB\sin 0^\circ \]
$\Rightarrow F = 0$
Also, we should notice that if the charged particle is stationary, then the force on the charged particle is zero as the velocity of the charged particle is zero that is v=0. This could be better illustrated by the above equation as,
\[ \Rightarrow F = q \times 0 \times B\sin \theta \]
\[\Rightarrow F = 0\]
$\therefore$ We can say that a stationary magnet does not interact with stationary charge as the charge being stationary would not experience any magnetic force and therefore it would not deflect. Hence, the option (D) is correct.
Note:
We should be clear in our mind that as the charge moves, it generally experiences the forces both due to the electric field and magnetic field. Now, this combined application force is known as the Lorentz force which is due to both the electric field and magnetic field on a charged particle.
Complete step by step solution:
We can recall that the magnetic force on a charged particle is orthogonal to the magnetic field and we can represent it by the formula as,
$ \Rightarrow F = qvB\sin \theta $
In the above formula, $B$ is the magnetic field vector, v is the velocity of the particle, and $\theta $ is the angle between the magnetic field and particle velocity. The Force due to the magnetic field on a moving charged particle is perpendicular to both the velocity plane and magnetic field plane.
Therefore, we can say that if the motion of the charged particle is parallel to the magnetic field then the force due to the magnetic field on a charged particle is zero. This can be illustrated by the above formula as,
\[ \Rightarrow F = qvB\sin 0^\circ \]
$\Rightarrow F = 0$
Also, we should notice that if the charged particle is stationary, then the force on the charged particle is zero as the velocity of the charged particle is zero that is v=0. This could be better illustrated by the above equation as,
\[ \Rightarrow F = q \times 0 \times B\sin \theta \]
\[\Rightarrow F = 0\]
$\therefore$ We can say that a stationary magnet does not interact with stationary charge as the charge being stationary would not experience any magnetic force and therefore it would not deflect. Hence, the option (D) is correct.
Note:
We should be clear in our mind that as the charge moves, it generally experiences the forces both due to the electric field and magnetic field. Now, this combined application force is known as the Lorentz force which is due to both the electric field and magnetic field on a charged particle.
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