An electron experiences a magnetic force of the form qv times B. If the magnetic

Question

An electron experiences a magnetic force of the form qv times B. If the magnetic field has magnitude 0.1 Tesla and points in the z direction, and the electron is traveling due north at half the speed of light, give me the numerical value of the z component of acceleration. Consider two vectors a = x + y + z and b = x - y. Give me a vector that is perpendicular to both of them. In problem 1.17 on page 36 of Taylor's mechanics book he states the cross product rule for derivatives: d/dt (r times s) = r times ds/dt + dr/dt times s Use that rule to work out the derivative of angular momentum (defined as L = r times p. Explain very carefully why the derivative of angular momentum only has one term, not two, even though the cross product rule for derivatives has two terms.

Explanation / Answer

solution (a) the force on the electron be

F = qvXB

= 1.6X10^-19X(3X10^8/2)X 0.1

Fz = 0.24X^-11 N

the acceleration on z component be = Fz / m ,where m is mass of the electron

az = 0.24X10^-11/ 9.1x10-31 ms-2

az = 0.0264x1020 ms-2

(b) for perprndicular vector, let us first consider three vectors A,B,and C

A = ai+bj+ck , B = xi +yj +zk and C = xi- yj

A = BXC

A = a^ X b^

= (xi+yj+zk) X (xi-yj)

=0 +xy k^ - zx j^ +xy k^- 0-yz i^

A = - ( xy i^ + zx j^ -2 xy k^ )

  

(C) as per Taylor mechanics   

d(r^ X s^) = rX ds/dt + dr^/dt X s^

angular momentum L = (r^ x p^)

dL/dt = d(r^Xp^) = dr/dt xp^ + dp/dt x r^

dL/dt = r X dp/dt ( since , dr/dt xp = vx mv = 0, wher v is velocity )

this is the reason that the derivative of angular momentum have only one term not two and dL/dt is also known as torque T and can be expressed as

T = dL/dt = r X F ( F = dp/dt )   

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