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An electron at pointAinFig. E27.15has a speedv0of1.41106m/sFind (a) the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path fromAtoB, and (b) the time required for the electron to move fromAtoB.

Short Answer

Expert verified
  1. B=1.610-4Tintothepage
  2. t=1.110-7s

Step by step solution

01

27.15 (a) - Direction

To find the direction we can apply the right hand rule as follows:

Define our initial velocity to be in the positive y direction, then our electron feels a force in the positive x direction. Thus our B field must be pointed into the page (recall the right hand rule is defined for positive charges).

02

27.15 (a) - Magnitude

To find the magnitude, we must use our formula for the magnetic force on a particle.

F鈬赌=qv鈬赌B鈬赌F=qvBsin()

The magnitude of the magnetic field is not changeing, so we can choose a point where it is most convenient to calculate the magnitude. Thus

F=qv0B

recall from Newton's second law that

F=ma

Where in this case, the acceleration is the centripital acceleration, so

a=v02r

Putting it all together we get

F=qv0Bma=qv0Bmv02r=qv0BB=mv0qr

These are all known values.

charge of an electron: q=-1.610-19C

mass of an electron: m=9.1110-31kg

The radius is half of our given diameter, so r=5.0cm=0.05m

Now we can plug in our values and evaluate.

B=(9.1110-31kg)(1.41106m/s)(1.6010-19C)(0.05m)

B=1.6*10-4kgCsB=1.610-4T

03

27.15 (b) - time

To find the time we can use rotational motion. The period is the time to complete one full circle, so our time is half the period

t=12Tt=122t=

Recall that our angular velocity, , is given in terms of the tangential velocity as follows:

=vr

Thus we have

t=t=v0rt=蟺谤v0

Plug in our numbers and evaluate

t=0.05m1.41106m/st=1.110-7s

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