contestada

A positively charged particle of mass 6.81 × 10-8 kg is traveling due east with a speed of 62.0 m/s and enters a 0.477-T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 2.83 × 10-3 s, at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge.

Respuesta :

Rau7star

Answer:

a) 2.34x [tex]10^-^3[/tex]N

b) 7.91 x [tex]10^-^5[/tex] C

Explanation:

a)The centripetal force is given by

F= mv²/r

For radius, using the equation i.e s=[tex]\frac{1}{4}[/tex] (2πr) and substituting s=vt there gives,

r=[tex]\frac{2s}{\pi }[/tex] =>[tex]\frac{2vt}{\pi }[/tex]

Now, using the equation for magnitude of magnetic force

F= mv²/r => mv²/ [tex]\frac{2vt}{\pi }[/tex]

F= [(3.142)(6.81 x [tex]10^-^8[/tex]kg)(62m/s)] / (2)(2.83x [tex]10^-^3[/tex]s)

F= 2.34x [tex]10^-^3[/tex]N

b) In order to find the magnitude |q| of the charge, we'll use the equation

F= |q|vBsinθ

|q|= F/vBsin(90°) => 2.34x [tex]10^-^3[/tex]/(62 x 0.477 x 1)

|q|= 7.91 x [tex]10^-^5[/tex] C

Otras preguntas