A cyclotron is opened at an oscillator frequency of 12 MHz and has a dee radius R = 50 cm. What is the magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron?
Answers
A circular current loop of magnetic moment M is in an arbitrary orientation in an external magnetic field B. The work done to rotate the loop by 30 about an axis perpendicular to its plane is Zero
Solution
Potential energy of a loop in a uniform magnetic field is given by
U=-MBcosθ
where,
M is magnetic moment of the current carrying loop,
B is magnetic field in which the loop is kept,
θ is the angle between M and B.
The change in potential energy of the loop will give the work done by the magnetic field as initial and final kinetic energy is 0. Even though the coil is rotated, about an axis perpendicular to its plane, the potential energy does not change. Since, there is no change in angle between M and B when we are rotating the loop along the axis perpendicular to its plane so, there will be no change in the potential energy of the loop.Hence, work done is zero.
Answer:
In optics, dispersion is the phenomenon in which the phase velocity of a wave depends on its frequency. Media having this common property may be termed dispersive media. Sometimes the term chromatic dispersion is used for specificity.