GREPhysics.NET: GR0177 #58

GR0177 #58

Problem

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Electromagnetism $\Rightarrow$ }Trajectory

The velocity of the proton as it enters the region is given by its acceleration due to the potential difference V. Conservation of energy yields $1/2 mv^2 =qV \Rightarrow v^2 = 2qV/m$ .

In the first experiment, since the electric and magnetic forces balance, as the particle is un-deflected, one deduces that $qvB=qE \Rightarrow vB = E$ from the Lorentz force.

In the second experiment, the potential is doubled to 2V. The velocity is thus greater than it was in the first experiment. Since the magnetic force is proportional to the velocity, one has the magnetic force larger. Thus, the particle is deflected in the direction of the magnetic field, which is in the -x-direction from the right hand rule, as in choice (B).

Alternate Solutions

BerkeleyEric
2010-07-18 19:23:00

Basically the electric field pushes the proton in +x and the magnetic field pushes it in -x, so when the velocity is increased (by increasing V), the magnetic field wins out.

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Comments

BerkeleyEric 2010-07-18 19:23:00	Basically the electric field pushes the proton in +x and the magnetic field pushes it in -x, so when the velocity is increased (by increasing V), the magnetic field wins out. Reply to this comment
niux 2009-11-02 13:16:46	Although magnetic field is now bigger, the particle is not deflected in the direction of magnetic field (+y), but in the direction resultant from vectorial product: v cross B (+z cross +y) i.e. in (-x) direction, then we have choice B. Reply to this comment
chahah 2009-11-01 10:25:37	The initial calculation could just be omitted, since the answer choices do not contain any numbers. Reply to this comment

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