A charged particle with a charge-to-mass ratio of |q|/m = 5.7 × 108 C/kg travels on a circular path that is perpendicular to a magnetic field whose magnitude is 0.84 T. How much time does it take for the particle to complete one revolution?

Time to complete one revolution = 1.31 x 10^(-8) s

Explanation:

In motion of charges in electromagnetic fields, we know that ;

R = mv/qB

Where,

R is radius of the circular path

q is the charge on the particle

m is the mass of the particle moving v is constant speed

B is magnetic field

Now in the question, we are given value of q/m. Let's rearrange the equation to show that;

r = mv/qB

(q/m)•(rB) = v - - - - (1)

Now, in circular motion, we know that; Period; T = 2πr/v

Thus, let's make v the subject.

v = 2πr/T - - - - - (2)

Now equating eq 1 to eq 2,we obtain;

(q/m)•(rB) = 2πr/T

r will cancel out to give ;

(q/m)B = 2π/T

Making T the subject, we get;

T = 2π/[(q/m)B]

From the question,

B = 0.84 T

q/m = 5.7 × 10^(8) C/kg

Thus,

T = 2π/[5.7 × 10^(8) x 0.84] = 1.31 x 10^(-8) s

massxgravity x heght=weightxheight= pe

4900=490xh

h=10