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protons having a kinetic energy of 5.20 mev are moving in the positive x-direction and enter a magnetic field of 0.0550 t in the z-direction, out of the plane of the page, and extending from x. 0→x=1m as shown in Fig.
a. Calculate the y-component of the protons' momentum as they leave the magnetic field.
b. Find the angle ϕ between the initial velocity vector of the proton beam and the velocity vector after the beam emerges from the field. Ignore relativistic effects and note that 1eV=1.60×10−19J.

Sagot :

a)  The y-component of the protons' momentum as they leave the magnetic field is 8.8 x 10-21 kg m/s

b) The angle ϕ between the initial velocity vector of the proton beam and the velocity vector after the beam emerges from the field is  9.812821°

Since are given with the kinetic energy of  5.20 Mev (8.32X 10^-13J), and  a magnetic field of 0.0550 t in the z-direction.From the diagram we can see  that, p(Y) =mv sin θ

=> sin θ =1/R , where R = mv/eB

=>sin θ = eB/mv

=>eB= mv sin θ

=> Py(the momentum) = eB =1.60×10−19J*0.0550 =  8.8 x 10-21 kg m/s .In the second case, sin θ  =  eB/mv, since the kinetic energy is 5.20 Mev

=>1/2 mv² =  5.20 mev = 8.33132e⁻¹⁹

=> v = √(2* 8.33132e-19 / 1.6 x10⁻²¹) = 32.27 m/s²

Now sin φ = eB/mv

=>sin φ =  8.8 x 10-21/ 1.6 x10^-21* 32.27 = 0.17043

=>φ = sin-1(0.17043)  = 9.812821°

To know more about momentum refer to the link brainly.com/question/24030570

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