Genius is one percent inspiration and ninety-nine percent perspiration.

– Thomas Alva Edison

Today I give you a** **free response practice question involving magnetic force on moving charges. The question is meant for AP Physics C aspirants. You may work out the question yourself and compare your answer with the model answer given below the question. Here is the question:

A proton (mass 1.67×10^{–27} kg, charge 1.6×10^{–19} C) at rest at the origin is acted on by a uniform electric field of intensity 900 V/m, acting along the positive X-direction in the region from x = 0 to x = 8.35 cm. A uniform magnetic field is applied perpendicular to the direction of motion of the proton in the region from x = 8.35 cm to x = 16.7 cm so that the proton is finally brought to a detector located at x = 0, y = – 8.35 cm, z = 0. Now answer the following questions:

**(a)** In which direction should the magnetic field be applied. Pick out the answer by putting a tick (√) mark against the correct option among the following:

(i) Along the negative Y-direction _____

(ii) Along the negative Z-direction _____

(iii) Along the positive Z-direction _____

Justify your answer.

**(b)** Calculate the velocity with which the proton enters the magnetic field.

**(c)** Calculate the magnetic flux density needed to bring the proton exactly at the detector at the point (0, – 8.35 cm, 0).

**(d) **The magnetic field is gradually decreased from the value asked for in part (c). Calculate the value of the magnetic field which will allow the proton to *just cross* the magnetic field. ** **

**(a**) The magnetic field is to be applied along the positive Z-direction. [Option (iii)].

At the moment the proton enters the magnetic field, its velocity is along the positive X-direction. The magnetic force on the electron at that moment should be downwards (along the negative Y-direction) so that it suffers a negative Y-displacement by the time it reaches the detector. By Fleming’s left hand rule (motor rule), this can be achieved only if the magnetic field is directed along the positive Z-direction.

**(b)** In the adjoining figure O is the origin from where the proton is released and P is the position of the detector. It is accelerated along the positive X-direction by the electric field *E* (let us say). The work done by the electric field on the proton in moving through a distance *d* along the electric field is *qEd* where *q* is the charge on the proton. This is equal to the kinetic energy of the proton at the moment it reaches the magnetic field. Therefore, we have

*qEd= *½ *mv*^{2} where *m *is the mass of the proton and *v* is its velocity.

Therefore, *v = *√(2*qEd/m)*

Or, *v = *√[(2×1.6×10^{–19}×900×8.35×10^{–2})^{ }*/*(1.67×10^{–27}*)*] = 1.2×10^{5} ms^{–1}, *directed along the positive X-direction.*

**(c)** In the magnetic field the proton moves along a circular path of radius *r* obtained by equating the magnetic force on the proton to the centripetal force:

*qvB = mv*^{2}/*r* where *B* is the magnetic flux density.

Therefore, *r = mv/qB* and the magnetic flux density needed to bring the proton exactly at the detector is given by

*B* = *mv/qr*

As the proton has to reach the detector at the point (0, – 8.35 cm, 0), *the diameter* of its semicircular path (as demanded by geometry) in the magnetic field is 8.35 cm so that the radius of the path is *r =* 4.175 cm.* *

Therefore, *B =* (1.67×10^{–27}×1.2×10^{5})/(1.6×10^{–19}×4.175×10^{–2}) = **3×1****0**^{–2}** tesla**.

**(d) **If the magnetic field is decreased from the above value, the radius of its path is increased. The limiting value of the magnetic field upto which the proton is prevented from crossing the field is that which will make the radius of the path equal to the extent of the magnetic field (*r* = 8.35 cm). The path of the proton in this case is indicated by the dotted semicircle in the figure.

Since the proton will cross the magnetic field on decreasing the field by a very small amount from this limiting value, it is the magnetic field which will allow the proton to *just cross* the magnetic field. This field (*B*_{1} )is given by

*B*_{1} = *mv/qr *where *r* = 8.35 cm = 0.0835 m

The value of *B*_{1} is *half *the value obtained in part (c) since the radius is doubled.

Therefore *B*_{1} = **1.5×1****0**^{–2}** tesla**.

## No comments:

## Post a Comment