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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

1 Answer

2 votes

Answer:

(A) q/m = 9.58 x 10⁷ C/kg

(B) b. Proton

Step-by-step 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

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