An electron moving with a speed of 1.50*10^7 m/sec is projected at right angles into a uniform magnetic field of flux density 6.50*10^-8 w/m^2. Calculate quantitatively the new path of the electron.

Respuesta :

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]