Question

PROBLEM 20.15: A deuteron particle (the nucleus of an isotope of hydrogen consisting of one proton and one neutron and having a mass of 3.34×10−27 kg ) moving horizontally enters a uniform, vertical, 0.700 T magnetic field and follows a circular arc of radius 56.5 cm .

How fast was this deuteron moving just before it entered the magnetic field ?

How fast was this deuteron moving just after it came out of the field?

What would be the radius of the arc followed by a proton that entered the field with the same velocity as the deuteron?

Answer 1

here m = 3.34 x 10^-27 Kg

this partical enter horizonal in the vertical uniform magnetic field

B = 0.700 T

here velocity of partical and magnetic field are in perpendicular direction so it follow circular motion

radius of circle = 56.5cm = 0.565 m

so as we know that

r = mv / qB

so the entered velocity

v = qrB / m

= ( 1.6 x 10^-19 x 0.565 x 0.700 ) / ( 3.34 x 10^-27 )

= 0.1894 x 10⁸ = 18.94 x 10⁶ m/s Ans

in the circular path the magnitude will constant but the direction will change

so the partical came out of the field of the same velocity

v_o = 18.94 x 10⁶ m/s Ans

if we use proton as same velocity

so as we know that

r = mv / qB

m = mass of proton = 1.672 x 10^-27 Kg

r = ( 1.672 x 10^-27 x 18.94 x 10⁶ ) / ( 1.6 x 10^-19 x 0.700 )

= 28,2747 x 10^-2 = 0.2827 m = 28.27 cm Ans