a) An electron enter a magnetic field. The magnetic field has a uniform strength

Question

a) An electron enter a magnetic field. The magnetic field has a uniform strength 2.4 T and is oriented at a 45 degree angle to the electron's motion. A force of magnitude 9.2810^-13 N acts on the electron. What is the electron's speed?

b) An alpha particle, which is a helium nucleus, has a mass of 6.64*10^-27 kg. It is travleing horizontally at 42.3 km/s when it enters a uniform, vertical 1.6 T magnetic field. What is the diameter of the path followed by the alpha particle?

c) An alpha particle, which is a helium nucleus, has a mass of 6.64*10^-27 kg. It is travleing horizontally at 42.3 km/s when it enters a uniform, vertical 1.6 T magnetic field. What is the magnitude of the acceleration of the alpha particle while it is in the magnetic field?

Explanation / Answer

of the form F=q(vxB)=qvBsintheta

Since the direction of velocity of the alpha particle is perpendicular to the applied magnetic field theta=90 thus sin theta=1

So F=qvB

Now due to the magnetic field the alpha particle will be deflected by a centripetal forceof the form F=mv^2/r

thus qvB=mv^2/r or r=mv/qB

thus r=(6.6*10^-27*1.6*10^7)/(2*1.6*10^-19*1.%u2026 <- since e=1.6*10^-19 C put 2e=2*1.6*10^-19

1) thus r=0.24 m=24 cm and the diameter of the path d=2r=48 cm

2) If the magnetic field were doubled then the new diameter is halved since r= 12 cm then and d=2r=24 cm (using the above formula again)

3) If the speed of the alpha particle is doubled then the new diameter is doubled. r=48 cm for v=2v and thus d=2r=96 cm.

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