Sagot :
Given;
The charge q=3 μC
The distance of the charge from the wire, r=1.5 cm=1.5×10⁻² m
The current through the wire, I=5 A
The velocity of the charge, v=1000 m/s
To find:
The magnitude and direction of the magnetic force.
Explanation:
The magnetic field at a distance r from the wire produced due to the current through the wire is given by,
[tex]B=\frac{\mu_0I}{2\pi r}[/tex]Where μ₀ is the permeability of free space.
On substituting the known values,
[tex]\begin{gathered} B=\frac{4\pi\times10^{-7}\times5}{2\pi\times1.5\times10^{-2}} \\ =6.67\times10^{-5}\text{ T} \end{gathered}[/tex]The direction of the magnetic field is given by the right-hand thumb rule and the magnetic field is perpendicular to the wire and the direction of motion of the particle.
The magnetic force acting on the charged particle due to this magnetic field is given by,
[tex]F=\text{qvB}[/tex]On substituting the known values,
[tex]\begin{gathered} F=3\times10^{-6}\times1000\times6.67\times10^{-5} \\ =2\times10^{-7}\text{ N} \end{gathered}[/tex]The direction of this force is given by the right-hand rule, according to which if the thumb is pointed in the direction of motion of the positive charge and force finger is pointed in the direction of the magnetic field, then the middle finger will point towards the magnetic force. Thus according to this rule, the force acting on the given charge is parallel to, and in the direction of the current flow.
Final answer:
The force acting on the charge is 2×10⁻⁷ N, parallel to and in the direction of the current flow.
Therefore the correct answer is option C.