Answer:
Explanation:
In a scenario where a particle of charge overrightarrow{B} enters a magnetic field with a velocity overrightarrow{V}, it experiences a force overrightarrow{F} given by: overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{B}).
implies F=BqVSin\theta.
Where theta is the angle between the velocity vector of the particle and the magnetic field vector.
When a particle enters the magnetic field at an angle 90, it moves in a circular path as it experiences a centripetal force, given by F=\frac{mV^2}{R}.
Where R is the radius of the circle, V is its velocity and m is its mass
Thus, magnetic force becomes F=BqVSin90^o=\frac{mV^2{R}\implies R=\frac{mV}{Bq}.
The equation changes as below, when velocity is doubled, let us assume that the radius is given by R_1.
R_1=\frac{2mV}{Bq}=2R.
Therefore, it is obvious that the velocity of a charged particle in a circular arc is directly proportional to the radius of the arc. The radius of the circular arc doubles when the velocity of the charged particle in the circular orbit doubles only if the mass, charge and magnetic field of the particle remains constant
Hence when velocity is doubled radius of the circle also gets doubled.
When the speed of the particle doubles, the radius of the arc also doubles.
The magnetic force on the particle is calculated as follows;
[tex]F = qvB[/tex]
The centripetal force on the particle is calculated as follows;
[tex]F_c = \frac{mv^2}{R}[/tex]
The speed of the particle is calculated as follows;
[tex]\frac{mv^2}{R} = qvB\\\\mv = qBR\\\\v = \frac{qBR}{m} \\\\\frac{v_1}{R_1} = \frac{v_2}{R_2}[/tex]
when the speed of the particle doubles;
[tex]\frac{v_1}{R_1} = \frac{2v_1}{R_2} \\\\R_2v_1 = 2R_1v_1\\\\R_2 = 2R_1[/tex]
Thus, we can conclude that when the speed of the particle doubles, the radius of the arc also doubles.
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