“There is only one corner of the universe that you can be certain of
improving, and that’s your own self.”

– Aldous Huxley

Many questions on electromagnetic induction have
been discussed on this site earlier. You can access those questions by clicking
on the label, ‘electromagnetic induction’ below this post. Today we shall discuss
a few more questions in this section.

(1) An object of mass *M* (Fig.) carrying a charge +*Q*
is released from rest at the top of a smooth inclined plane of inclination *θ*.*
*A magnetic field of flux density B acts throughout the region and is directed *normally into* the plane of the figure.
Which one among the following statements is correct?

(a) The acceleration of the object down the
incline depends on *Q* but does not
depend on B.

(b) The acceleration of the object down the incline depends on B but
does not depend on *Q*.

(c) The acceleration of the object down the
incline depends on both *Q* and B.

(d) The acceleration of the object down the
incline does not depend on *Q* and B.

(e) All the above statements are incorrect.

The component of the gravitational force along
the incline is *Mg*sin*θ* which can produce in the object an
acceleration *g*sin*θ *down the incline. When the object moves down along the inclined
plane, the magnetic force (Lorentz force) acting on the object is directed
perpendicular to the inclined plane such that it tries to press the object
against the inclined plane.

[Apply Fleming’s left hand rule to arrive at the
direction of the magnetic Lorentz force].

Since the magnetic force on the object is
perpendicular to the inclined plane, it has no component along the plane. The
acceleration of the object down the incline is *g*sin*θ* itself and is
independent of *Q* and B [Option (d)].

(2) In the above question suppose the inclined
plane is rough and the object is able to slide down. Which one among the
following statements is correct?

(a) The acceleration of the object down the
incline depends on *Q* but does not
depend on B.

(b) The acceleration of the object down the
incline depends on B but does not depend on *Q*.

(c) The acceleration of the object down the
incline depends on both *Q* and B.

(d) The acceleration of the object down the incline
does not depend on *Q* and B.

(e) All the above statements are incorrect.

In this case the frictional force acting between
the object and the inclined plane reduces the acceleration of the object. The
frictional force *F *depends on the
normal force *N *acting on the object.

[Remember that *F =* *μN
*where *μ *is the coefficient of friction]

But the normal force (which is *Mg*cos*θ* + magnetic Lorentz force) depends on
both *Q* and B since the magnetic
Lorentz force is *QvB* where *v *is the velocity of the object.

Therefore, the acceleration of the object down
the incline depends on both *Q* and B
[Option (c)].

(3) A plane rectangular coil of length 2*L* and breadth *L* has *N* turns of
insulated copper wire in it. Initially (at time *t* = 0) the plane of the coil is perpendicular to a uniform magnetic
field of flux density *B* . The coil
rotates with angular velocity ω such that the axis of rotation of the coil is
at right angles to the magnetic field (Fig.). If the voltage induced in the
coil is *V*_{max}sin*ωt*, the angular velocity of the coil is

(a) *V*_{max
}/ 2*NL*^{2}*B*

(b) 2*V*_{max
}*t* / *NL*^{2}*B*

(c)*V*_{max}/*NL*^{2}*B *

(d) *NL*^{2}*B/ V*_{max}

(e)* *2*NL*^{2}*B/ V*_{max}
The magnetic flux *φ *linked with the
coil at the instant *t* is given by

φ = *NAB *cos
*ωt* where *A* is the area of the coil.

The voltage *V*
induced in the coil is is given by

*V *= – *d**φ **/dt*
= *NABω *sin *ωt*

It is given in the question that *V *=* **V*_{max}sin*ωt*

Therefore, *V*_{max}
= *NABω*

This gives *ω
= V*_{max }/*NAB *

Since the area *A* = 2*L*^{2}, we
have* ω = V*_{max }/ 2*NL*^{2}*B*

(4) In the transformer shown in the adjoining
figure, the secondary winding has taps A, B, C, D and E. The number of turns between
taps A and B is 100. The number of turns between taps B and C is 50, the
number of turns between C and D is 40 and that between D and E is 10. The
primary has 2000 turns and is connected to 220 volt A. C. mains. Between which
taps will you obtain 5.5 volt output?

(a) Between A and B

(b) Between A and B as well as B and E

(c) Between B and C as well as C and E

(d) Between C and D

(e) Between D and E

Since the primary voltage is 220 V and the number
of turns in the primary is 2000, the induced voltage per turn is 220/2000 =
0.11 volt.

Therefore, the number of turns required to
produce a voltage of 5.5 volt is (5.5)/(0.11) = 50.

Since there are 50 turns between taps B and C as
well as between taps C and E, the correct option is (c).

[You can work this out this way also:

Considering the entire secondary of the
transformer, the secondary to primary turns ratio is 200/2000 = 1/10.
Therefore, the secondary voltage (between the ends A and E) is 220/10 = 22 V.
Since the voltage across the entire secondary containing 200 turns is 22 volt,
we will get 5.5 volt across 50 turns (between taps B and C as well as between
taps C and E)]