A 5 g particle carrying a charge of 76 mu C enters a uniform 6 T magnetic field

Question

A 5 g particle carrying a charge of 76 mu C enters a uniform 6 T magnetic field at a speed of 44 m/s and with an angle of 50 degree with respect to the field lines, as shown in the figure. For each of the following statements, determine whether it is true or false. The y-component of the particle's velocity is unchanged as it passes through the B-Field. The Force on the particle is in the -z direction. The field does a finite amount of work on the particle as the particle's trajectory is bent by the field. The particle's speed is unchanged as it passes through the B-Field. The particle follows a circular path as the force is normal to the velocity.

Explanation / Answer

[Underscore is used here to denote a vector quantity.]
the Lorentz force law:

F_ = q(E_ + v_ x B_)

Since E_ = 0 here, that's just
F_ = qv_ x B_

and this is true whatever the angle between v_ and B_. In any cross product,
the result is both vectors, so
F_ v_, i.e., F_•v_ = 0,
and so no work is done by the field on the particle.
A) is False

B) is True initially, because v_ is in the xy-plane, with a +y-component, and so
v_ x B_ has only a (+y) x (+x) = (-z)-component

But during the particle's travel, the force on it rotates in the yz-plane, and doesn't stay in the -z direction
C) is True

And because
F_ B_
and B_ || x, the force is x, and so the x-acceleration is 0, and



D) is True because of A) -- no work done = no change in kinetic energy = no change in speed

E) is False ,.

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