The given situation is shown below.
Given,
v=1000m∕ sq=10−16Cm=1×10−27kg N= Number of turns
=5000I=5AR= Radius of circular path of charge
Magnetic field,
B=µ0nI (where,
n= number of turns per unit length)
⇒B=µ0I(where,
n= )
⇒B=µ0××5⇒B=µ0×........(i)
Also we know that, Distance
= speed
× time
L=vx×t (where,
vx is the component of velocity along the axis of the solenoid)
⇒L=500t......(ii)
Also, radius of circular path of a charged particle in uniform magnetic field is given by the formula
R=(. where,
vy=500√3) Putting,
B=[from Eq .(i)]
⇒R=......(ii)
Now, time taken by the charged particle to exit the solenoid moving along the circular path making
N′ number of revolutions.
t′=⇒t′=| 2πR× no. of revolutions |
| vy |
t′= t′=| 2π×10−11×√3L×N′ |
| 50µ0×500√3 |
.....(iii)
Now, since time taken to cover linear distance and circular distance before exiting the solenoid are same
t=t′ So, from Eqs. (i) and (ii)
=| 2π×10−11×√3×L×N′ |
| 50×4π×10−7×500√3 |
(∵µ0=4π×10−7) On solving, we get
N′===106 Hence, total number of revolution along helical path by the time it emerges the solenoid
=1×106. Hence, option (b) is the correct answer.
