AP EAPCET 06-Jul-2022 Shift 2 Paper

The given situation is shown below.Given, v=1000m∕ s
‌q=10−16C
‌m=1×10−27‌kg
N= Number of turns =5000
I=5A
R= Radius of circular path of charge
Magnetic field, B=µ0nI (where, n= number of turns per unit length)
⇒B=µ0‌

N

L

I
(where, n=‌

N

L

)
‌⇒B=µ0×‌

5000

L

×5
‌⇒B=µ0×‌

25000

L

........(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=‌

mvy

9B

(. where, vy=500√3)
Putting, B=‌

µ0×25000

L

[from Eq .(i)]
⇒‌‌R=‌

10−11×√3L

50µ0

......(ii)
Now, time taken by the charged particle to exit the solenoid moving along the circular path making N′ number of revolutions.
‌t′=‌

‌ circular distance ‌

vy

⇒t′=‌

2πR×‌ no. of revolutions ‌

vy

‌t′=‌

2πR×N′

vy

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)
‌

L

500

=‌

2π×10−11×√3×L×N′

50×4π×10−7×500√3

‌‌(∵µ0=4π×10−7)
On solving, we get
N′=‌

100×10−7

10−11

=‌

10−5

10−11

=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.