In Fig. 28-30, a charged particle enters a uniform magnetic field with speed , moves through a halfcirclein time , and then leaves the field
. (a) Is the charge positive or negative?
(b) Is the final speed of the particle greater than, less than, or equal to ?
(c) If the initial speed had been , would the time spent in field have been greater than, less than, or equal to ?
(d) Would the path have been ahalf-circle, more than a half-circle, or less than a half-circle?
Consider the formula for the radius of the charge as:
Here, r is radius, B is magnetic field, v is velocity, m is mass,q is charge on particle
The charge on the particle:
In order to get the force on the particle downwardaccording to the right hand rule, the charge on the particle must be negative.
Hence, the charge on the particle is negative.
The final speed of the particle greater than, less than, or equal to :
Magnetic field does not do any work on the charge particle. It only changes the direction of velocity in magnetic field region.
Hence, the speed of the particle does not change.
Hence, its final speed is the same as the initial speed.
If the initial speed is , then whether the time spent in the magnetic field is greater than, less than, or equal to :
If velocity is then using the above relation,
Hence, radius will get halved.
Consider the formula:
Since the speed is halved and the radius is halved, the period will not change. It will be the same as the initial time
Hence, if the initial speed is , then the time spent in the magnetic field is equal to
Path is half circle, more than half circle, or less than a half circle:
The path will be half circle. Speed does not change the direction; only the magnetic force changes the direction.
Hence, the path would have been a half circle.
Therefore, use the right hand rule to find the force on the particle and usethe relation of radius and magnetic field to find the radius of different particles.
Fig. 28-49 shows a current loop ABCDEFA carrying a current i = 5.00 A. The sides of the loop are parallel to the coordinate axes shown, with AB = 20.0 cm, BC = 30.0 cm, and FA = 10.0 cm. In unit-vector notation, what is the magnetic dipole moment of this loop? (Hint: Imagine equal and opposite currents i in the line segment AD; then treat the two rectangular loops ABCDA and ADEFA.)
Figure 28-52 gives the orientation energy U of a magnetic dipole in an external magnetic field , as a function of angle between the directions , of and the dipole moment. The vertical axis scale is set by . The dipole can be rotated about an axle with negligible friction in order that to change . Counterclockwise rotation from yields positive values of , and clockwise rotations yield negative values. The dipole is to be released at angle with a rotational kinetic energy of , so that it rotates counterclockwise. To what maximum value of will it rotate? (What value is the turning point in the potential well of Fig 28-52?)
Question: An electron has velocity as it enters a uniform magnetic field What are (a) the radius of the helical path taken by the electron and (b) the pitch of that path? (c) To an observer looking into the magnetic field region from the entrance point of the electron, does the electron spiral clockwise or counterclockwise as it moves?
Figure 28-24 shows a metallic, rectangular solid that is to move at a certain speed v through the uniform magnetic field . The dimensions of the solid are multiples of d, as shown.You have six choices for the direction of the velocity: parallel to x, y, or z in either the positive or negative direction.
(a) Rank the six choices according to the potential difference set up across the solid, greatest first.
(b) For which choiceis the front face at lower potential?
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