A charged particle travels along a straight line with a speed $\nu$ in a region where both electric field $\mathbf{E}$ and magnetic fields $\mathbf{B}$ are present. It follows that}

Question

The pattern of the magnetic field produced inside a current carrying solenoid is :

Answer 1

A charged particle moving in a straight line within a region containing both electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$ presents us with an interesting scenario. We know that the force acting on a charged particle in the presence of electric and magnetic fields is given by the Lorentz force equation:

$\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$

Where:

$\mathbf{F}$ is the total force on the particle.
$q$ is the charge of the particle.
$\mathbf{v}$ is the velocity of the particle.
$\mathbf{E}$ is the electric field.
$\mathbf{B}$ is the magnetic field.

For the particle to travel in a straight line, the net force $\mathbf{F}$ acting on it must be zero. Thus, we have:

$q(\mathbf{E} + \mathbf{v} \times \mathbf{B}) = 0$

Since $q \neq 0$ (as it is a charged particle), we simplify to:

$\mathbf{E} + \mathbf{v} \times \mathbf{B} = 0$

Rearranging gives:

$\mathbf{E} = -(\mathbf{v} \times \mathbf{B})$

This indicates that the electric field $\mathbf{E}$ is equal and opposite to the vector cross product of the velocity $\mathbf{v}$ and magnetic field $\mathbf{B}$.

For the cross product $\mathbf{v} \times \mathbf{B}$ to be non-zero, $\mathbf{v}$ and $\mathbf{B}$ must be perpendicular to each other.

Hence, the correct option that satisfies these conditions is:

$\mathbf{E} = \nu \mid \mathbf{B}$ and the two fields are perpendicular

A charged particle travels along a straight line with a speed $\nu$ in a region where both electric field $\mathbf{E}$ and magnetic fields $\mathbf{B}$ are present. It follows that}

Show Hint

The Correct Option is B

Solution and Explanation

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