Question

A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpendicular to the proton's velocity. It leaves the field-filled region with velocity -20.0j Mm/s. Determine(b) the radius of curvature of the proton's path while in the field.

Answer 1

The radius of curvature of the proton's path while in the field is $66.67$  × $10^{-2}$.

b) Let R = radius curvature of protons path. Then,

relation b/w B, R, and v is: -

$B = mv/eR\\R=mv/eB$

$R=\frac{1.6*10^{-27} * 20*10^{6}}{1.6*10^{-19}*0.3 }$

$R =66.67$× $10^{-2}$

Hence, the radius of curvature of the proton's path while in the field is $66.67$ × $10^{-2}$.

What do you mean by Magnetic field?

The magnetic influence on moving electric charges, electric currents and magnetic materials is described by a magnetic field, which is a vector field. A force perpendicular to the charge's own velocity and the magnetic field acts on it when the charge is travelling through a magnetic field.  The magnetic field of a permanent magnet pulls on ferromagnetic substances like iron and attracts or repels other magnets. A magnetic field that varies with location will also exert a force on a variety of non-magnetic materials by changing the velocity of those particles' outer electrons. Electric currents, like those utilized in electromagnets, and electric fields that change in time produce magnetic fields that surround magnetized things.

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