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 the electric field but the magnetic field continues, and the particle makes a semicircle of diameter 25.05 cm. Part A. What is the particles charge to mass ratio?Part B. Can you identify the particle? a. can't identify
b. proton
c. electron
d. neutron

(A) q/m = 9.58 x 10⁷ C/kg

(B) b. Proton

Explanation:

(A)

In order to pass un-deflected between magnetic and electric field both electric and magnetic forces must be equal:

Felectric = Fmagnetic

Eq = Bqv

v = E/B   equation (1)

Now, when the particle forms semi-circle under influence of magnetic field. At this, point magnetic field is equal to centripetal force:

Centripetal Force = Magnetic Force

mv²/r = qvB

mv = qrB

v = qrB/m   equation (2)

comparing equation (1) and equation (2), we get:

E/B = qrB/m

q/m = E/B²r

where,

q/m = charge to mass ratio = ?

E = Electric Field = 187500 V/m

B = Magnetic Field = 0.125 T

r = radius of circular path = 25.05 cm/2 = 12.525 cm = 0.12525 m

Therefore,

q/m = (187500 V/m)/(0.125 T)²(0.12525 m)

q/m = 9.58 x 10⁷ C/kg

(B)

This is the charge to mass ration of a proton:

q/m = 1.6 x 10⁻¹⁹ C/1.67 x 10⁻²⁷ kg

q/m =  9.58 x 10⁷ C/kg

Therefore, correct option is:

b. proton

answer; how far out of place which the object finished relative to the starting location;