An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and 0.1250 T, respectively. The particle passes out of the electric field, but the magnetic field continues, and the particle makes a semicircle of diameter 25.05 cm. Part A. What is the particle's charge-to-mass ratio?
Part B. Can you identify the particle?
a. can't identify
b. proton
c. electron
d. neutron

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic  fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is

The given formula to find the electric force is

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

Therefore,  the particle's charge-to-mass ratio is

b)

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

c) up is the direction the magnetic force is acting on the charge.

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm