A charged particle moves through a velocity selector at a constant speed in a straight line. The electric field of the velocity selector is 3.10 103 N/C, while the magnetic field is 0.360 T. When the electric field is turned off, the charged particle travels on a circular path whose radius is 4.20 cm. Find the charge-to-mass ratio of the particle.

The charge-to-mass ratio of the particle is 5.7 × 10⁵ C/kg

Explanation:

From the formulae

F = qvB and F = mv²/r

Where F is Force

q is charge

v is speed

B is magnetic field strength

m is mass

and r is radius

Then,

qvB = mv²/r

qB = mv/r

We can write that

q/m = v/rB (1)

Also

From Electric force formula

F = Eq

Where E is the electric field

and magnetic force formula

F = Bqv

Since, electric force = magnetic force

Then, Eq = Bqv

E = Bv

∴ v = E/B

Substitute v = E/B into equation (1)

q/m = (E/B)/rB

∴ q/m = E/rB²

(NOTE: q/m is the charge to mass ratio)

From the question,

E =  3.10 ×10³ N/C

r = 4.20 cm = 0.0420 m

B = 0.360 T

Hence,

q/m = 3.10 ×10³ / 0.0420 × (0.360)²

q/m = 569517.9306 C/kg

q/m = 5.7 × 10⁵ C/kg

Hence, the charge-to-mass ratio of the particle is 5.7 × 10⁵ C/kg.

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2