Question

A cylindrical solenoid has a length of 1- 160 cm. It is wound tightly with 4800 turns of wire, and the wire carries a current/-4.0 A. What is the period of cyclotron motion for a proton inside the solenoid? a. 4.35 b. 1.45 C 14 10 2 0.6.9 x 101

Answer 1

We have,
Length of the solenoid
1= 160 cm = 160 x 10 160 x 10-2 m

Turn of the solenoid
N = 4800

Current in the solenoid coil
+1 = 4 HA = 4 x 10-6 A

Permeability of space
→ Mo = 47 x 10-7 TmA-1

Mass of proton
→ m = 1.6726219 x 10-27 kg

Charge of the proton
19 q=1.6 x 10 с

The magnetic field inside the solenoid is given as:

B HON1


47 x 10-7 X 4800 x 4 x 10-6 ..BE 160 x 10-2

:: B = 1.508 x 10-8 1

The period of the proton inside the solenoid is given as:

2лт Т. qB

::T= 21 x 1.6726219 x 10 -27 1.6 x 10-19 X 1.508 x 10-8

:::T= 4.3557 s

Therefore, the period of the proton inside solenoid  
T= 4.3557 s
. Therefore, the correct option is ${\color{Blue} 4.3\ s}$ .