Magnetic Circular Motion!

Electricity and Magnetism Level 5

A particle mass $m$ charge $q$ is moving on a circular path on the surface of a friction less table with speed $v_o$ where a magnetic field $B_o$ exists uniformly over the whole region. It is attached by a string which passes through a hole in the table to a spring as shown. The spring is stretched by $x_o$ . If now the magnetic field is increased slowly to $2B_o$ ,

(1) The extension in the spring will increase.

(2) The speed of the particle will increase.

(3) The speed of the particle will decrease.

(4) The kinetic energy of the particle will decrease.

Note - If your answer is $(1,2)$ , Please mark your answer as $3 (=1+2)$

The answer is 7.

2 solutions

Lu Chee Ket
Dec 15, 2015

With 3 + 4 = 7 as answer, I try my best to explain accordingly:

The question doesn't eliminate ambiguity of direction of pull of the magnetic force generated! Therefore, we ought to guess for proper combinations from either way for a possible answer. Combination from answers suggest that the magnetic force is pulling it outward from center of rotation.

This seems no easy to explain as it involves both conservation of angular momentum and also the magnetic force perpendicular to both direction of motion and direction of flux which is proportional to speed of charge, particularly this involves with pulling spring instead of free electron in magnetic field. Conversion between K.E. and P.E. is of course not the less.

The consequence of the answer is potential energy increased while speed of the charged object decreased and therefore kinetic energy decreased but it is staying on the original orbit with no change of radius and no change of extension of spring. Conservation of energy must be complied for perpendicular forces acting onto moving object, and here, some K.E. is converted into P.E.

Initially, the charged particle tends to be pulled away from its orbit onto orbit of a greater radius. To conserve in angular momentum that $m r_1 v_1$ = $m r_2 v_2$ , the speed decreases. When the speed decreases, the magnetic force is also decreases, therefore it restores onto the original orbit again. This complies with the phrase of "the magnetic field is increased slowly to 2 $B_0$ ".

So, we have less speed for the circular motion while angular momentum conserved plus total energy conserved. An unseen out pull of magnetic force slow down the particle while giving it more potential energy onto exactly the same orbit eventually.

Experiment may help to verify if this is true.

U r making it complicated for no reason! Its very easy!! As u will change magnetic feild, flux will change as result, charge will experience a tangential force opposite to its motion ( from Lenz law, as B will increase, charge will try to oppose the change and hence it will experience duecto the induced electric field.) And hence its velocity and KE decreases. Yes i know problem says B is changed slowly, but if u ignore the induced E, it will lead to contradictions as there will be no tangential force, so no change in KE and if radius changes, then angular momentum will also change, but as there is no tangential force, there is no torque so its contradictory

Jatin Narde - 5 years, 4 months ago

Pranav Rao
Jan 19, 2016

I think the question wants us to ignore the induced electric field due to change in the magnetic flux. Otherwise there will be a torque on the moving charge about the center of Rotation due to force applied by this induced electric field. And so the angular momentum won't be conserved.

Magnetic Circular Motion!

The answer is 7.

2 solutions

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