Answer:
Explanation:
Given :
Speed of proton [tex]v = 3.2 \times 10^{6}\frac{m}{s}[/tex]
Mass of proton [tex]m = 1.67 \times 10^{-27}[/tex] Kg
The force on the proton in magnetic field is given by,
[tex]F = q (v \times B)[/tex]
[tex]F = qvB \sin \theta[/tex]
But [tex]\sin 90 = 1[/tex] (∵ Force is perpendicular to the velocity so [tex]\theta = 90[/tex])
[tex]F = qvB[/tex]
When particle enter in magnetic field at the angle of 90° so particle moves in circle
So force is given by,
[tex]F = \frac{mv^{2} }{r}[/tex]
Where [tex]r =[/tex] radius but in our case 0.23 m, [tex]q = 1.6 \times 10^{-19}[/tex] C
By comparing above two equation,
[tex]B = \frac{mv}{qr}[/tex]
[tex]B = \frac{1.67 \times 10^{-27} \times 3.2 \times 10^{6} }{1.6 \times 10^{-19} \times 0.23 }[/tex]
[tex]B = 0.145[/tex] T
