The following questions will be of common interest to students preparing for AP Physics B as well as AP Physics C examinations:

*R*carrying a steady current I is plotted against the distance of the point P from the centre of the wire. Which graph among the five options shown in the figure represents the correct plot?

Normally, you will consider the magnetic field due to thin straight conductors and you know that the field outside at distance *r *is given by

*B* =*µ*_{0}I/2π*r*

If you remember that the field is directly proportional to the current I and inversely proportional to the distance *r*, you can solve the above problem.

Since the field is inversely proportional to the distance, if the current is constant, the graph must be a *hyperbola*, as is the case for *points outside* the wire. In the case of points *within *the wire ( where r is less than the radius R)

*i* = I(πr^{2}/ πR^{2}) = I r^{2}/R^{2}

[We have assumed that the current is uniformly distributed across the section of the wire. In the case of *any point*, we need consider the contribution due to the current passing through a circular section of radius *r* only, remembering that the magnetic field within a current carrying pipe is zero everywhere. Try and prove it!]

In the case of points within the wire, the magnetic field is obtained by replacing the constant current I with the variable current *i *in the equation, *B* =*µ*_{0}I/2π*r.*

Thus, *B* = *µ*_{0} (I r^{2}/R^{2})^{ }/2π*r* = *µ*_{0}I r/ 2πR^{2}

The field *within* the wire is thus *directly proportional* to the distance *r.*

The correct plot is shown in figure (e).

(2) A free rectangular current carrying loop ABCD is placed as shown, near a long straight conductor PQ carrying a current I. The plane of the loop is the same as the plane containing the straight conductor and two sides of the loop are parallel to the straight conductor. The loop will

(a) move towards the straight conductor

(b) move away from the straight conductor

(c) rotate clockwise

(d) rotate anticlockwise

(e)** **have translational as well as rotational motion** **

Since PQ and DA carry *like currents* (currents in the same direction), they attract each other. You can easily show this by Fleming’s left hand rule (motor rule).. Since PQ and CB carry *unlike currents* (currents in opposite directions), they repel each other. The attractive force is greater than the repulsive force in this case since the separation between PQ and DA is greater than that between PQ and CB. [Remember that the force is directly proportional to the product of currents and inversely proportional to the separation]. The loop therefore will move towards the straight conductor.

At the loop, the magnetic field produced by PQ is directed perpendicularly into the plane of the figure. Therefore, the magnetic force on side AB is upwards and that on side CD is downwards. Since the magnitude is the same, the resultant of these two forces is zero The lines of action of these two forces are not separated . So, there is no torque. The lines of action of the forces on BC and AD also are not separated and again there is no torque.

The correct option therefore is (a)

(3) Two negative ions of the *same charge but different masses* are projected vertically upwards with the *same velocity* from point O (fig.) in a region of space where a uniform magnetic field directed perpendicularly in to the plane of the figure exists. If the gravitational pull is ignored, the trajectory of the heavier ion is

(b) OB

(c) OC

(d) OD

(e) OE

This is a question in which Flemings left hand rule (motor rule) will help you. It is better that you treat the ions to be positive (for the time being) so that you can hold the middle finger of your left hand along the direction of projection (upwards). The forefinger should indicate the direction of the magnetic field and hence must point into the plane of the figure. Then the direction of the thumb which is held perpendicular to the fore finger as well as the middle finger gives the direction of the magnetic force. The direction is leftwards for the positive ion. Since the ions in the question are negative, The deflection is towards right. Since the ions are of the same charge and the same velocity, the heavier ion is deflected less (because of greater inertia). Therefore its path is OD.

[If you remember the expression for radius *r, *which is

*r = mv/qB*, you will immediately note that *r* is greater for the larger mass *m*.

(4) In a mass spectrograph, positive ions moving with different velocities in the positive X-direction are admitted into a velocity filter in which an electric field of magnitude *E* is applied in the negative Y-direction. What should be the direction of the magnetic field so that ions of the same velocity pass through the velocity filter without deflection?

(a) Positive Z-direction

(b) Positive Y-direction

(c) Positive X-direction

(d) Negative Y-direction

(e)** **Negative Z-direction** **

As you might be knowing, the velocity filter is an arrangement in which electric and magnetic fields are applied at right angles so that the deflection produced by the electric field is cancelled by the deflection produced by the magnetic field. In other words, the electric force and the magnetic force on the ion are equal and opposite.

Since the electric field is in the negative Y-direction, the electric force (on the positive ion) is in the negative Y-direction. The *magnetic force* must therefore be in the *positive Y-direction. *

Now, to apply Fleming’s left hand rule, hold the thumb, fore finger and middle finger of your left hand in mutually perpendicular directions such that the middle finger is in the positive X-direction (to indicate the velocity of positive charge) and the *thumb* is in the* positive Y-direction* (to indicate the required direction of the* *magnetic force). The fore finger will then point along the required direction of the magnetic field. You will get it along the negative Z-direction

(5) In the above question, if the electric field is *E*, what should be the magnitude of the magnetic field so that ions of velocity *v* and charge *q* proceed undeflected through the velocity filter?

(a) *q/E *

(b) *Eq/v*

(c) *E/v*

(d) *v/E*

(e)** ***vq/E*** *** *

The electric force and the magnetic force on the ion must be equal and opposite. Equating the magnitudes, we have

*qE* = *qvB*

Therefore, *B *= *E*/*v*

You can expect more questions in this section in due course.

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