Answer:
the path of the moving charge will be circular path now
Radius = 1312.5 m
Explanation:
Force on a moving charge due to constant magnetic field is given by
[tex]\vec F = q(\vec v \times \vec B)[/tex]
since here force on the moving charge is always perpendicular to the velocity always as it is vector product of velocity and magnetic field so here magnitude of the speed is always constant
Also the force is since perpendicular to the velocity always
so here the path of the moving charge will be circular path now
now to find out the radius of this circular path
[tex]F = \frac{mv^2}{R}[/tex]
[tex]qvBsin90 = \frac{mv^2}{R}[/tex]
[tex]R = \frac{mv}{qB}[/tex]
[tex]R = \frac{9.1\times 10^{-31}(1.50 \times 10^7)}{1.6 \times 10^{-19}(6.50 \times 10^{-8})}[/tex]
[tex]R = 1312.5 m[/tex]