Answer:
1.312 x [tex]10^{-8}[/tex] s.
Explanation:
Given:
charge-to-mass ratio of "|q|/m" = 5.7 × 108 C/kg
magnetic field 'B'= 0.84T
We know that radius 'r' of the circular path of a charged particle 'q' of mass 'm', moving at constant speed 'v', in a uniform magnetic field 'B' can be defines as
r= mv / |q|B
so, v = (|q| B r)/m -->eq1
Now, also the linear velocity of a particle undergoing circular motion can be defined as
v= 2πr / t
so, t = 2πr / v
where,
2πr = total distance covered in one full revolution
t = time taken for one full revolution
Putting the value of v from eq 1 in above equation
therefore,
t= 2πr / ([tex]\frac{|q|}{m}[/tex] B r)
t= 2π / ( [tex]\frac{|q|}{m}[/tex]B )
t= 2π / (5.7 x [tex]10^{8}[/tex] x 0.84)
t= 1.312 x [tex]10^{-8}[/tex] s.
Therefore, it took 1.312 x [tex]10^{-8}[/tex] s for the particle to complete one revolution.
Answer:
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
