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#### Page No 229:

#### Question 6.1:

Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f ).

**(a)**

**(b)**

**(c)**

**(d)**

**(e)**

**(f)**

#### Answer:

The direction of the induced current in a closed loop is given by Lenz’s law. The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.

Using Lenz’s rule, the direction of the induced current in the given situations can be predicted as follows:

**(a)** The
direction of the induced current is along **qrpq***. *

**(b)** The
direction of the induced current is along **prqp**.

**(c)** The
direction of the induced current is along * yzxy*.

**(d)** The
direction of the induced current is along * zyxz*.

**(e)** The
direction of the induced current is along * xryx*.

**(f) **No current is induced since the field lines are lying in
the plane of the closed loop.

#### Page No 230:

#### Question 6.2:

Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19:

**(a) ** A wire of irregular shape turning into a circular shape;

**(b) ** A circular loop being deformed into a narrow straight wire.

#### Answer:

According to Lenz’s law, the direction of the induced *emf* is such that it tends to produce a current that opposes the change in the magnetic flux that produced it.

**(a) **When the shape of the wire changes, the flux piercing through the unit surface area increases. As a result, the induced current produces an opposing flux. Hence, the induced current flows along adcb.

**(b) **When the shape of a circular loop is deformed into a narrow straight wire, the flux piercing the surface decreases. Hence, the induced current flows along

#### Page No 230:

#### Question 6.3:

A
long solenoid with 15 turns per cm has a small loop of area 2.0 cm^{2
}placed inside the solenoid normal
to its axis. If the current carried by the solenoid changes steadily
from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop
while the current is changing?

#### Answer:

Number of turns on the solenoid = 15 turns/cm = 1500 turns/m

Number of turns per
unit length, *n *= 1500 turns

The solenoid has a
small loop of area, *A* = 2.0 cm^{2} = 2 × 10^{−4}
m^{2}

Current carried by the solenoid changes from 2 A to 4 A.

Change
in current in the solenoid, *di *= 4 − 2 = 2 A

Change in time, *dt*
= 0.1 s

Induced *emf* in
the solenoid is given by Faraday’s law as:

Where,

= Induced flux through the small loop

= *BA *...
(*ii*)

*B *= Magnetic
field

=

μ_{0} =
Permeability of free space

*= *4π×10^{−7}
H/m

Hence, equation (*i*)
reduces to:

Hence, the induced voltage in the loop is

#### Page No 230:

#### Question 6.4:

A
rectangular wire loop of sides 8 cm and 2 cm with a small cut is
moving out of a region of uniform magnetic field of magnitude 0.3 T
directed normal to the loop. What is the emf developed across the cut
if the velocity of the loop is 1 cm s^{−1}
in a direction normal to the (a) longer side, (b) shorter side of the
loop? For how long does the induced voltage last in each case?

#### Answer:

Length of the rectangular wire, *l* = 8 cm = 0.08 m

Width of the rectangular wire, *b* = 2 cm = 0.02 m

Hence, area of the rectangular loop,

*A* =* lb*

= 0.08 × 0.02

= 16 × 10^{−4}
m^{2}

Magnetic field strength, *B* = 0.3 T

Velocity of the loop, *v* = 1 cm/s = 0.01 m/s

**(a) **Emf developed in the loop is given as:

*e*
= *Blv*

= 0.3 × 0.08 ×
0.01 = 2.4 × 10^{−4}
V

Hence,
the induced voltage is 2.4 ×
10^{−4} V which lasts for 2 s.

**(b) **Emf developed,* e* = *Bbv*

= 0.3 × 0.02 ×
0.01 = 0.6 × 10^{−4}
V

Hence,
the induced voltage is 0.6 ×
10^{−4} V which lasts for 8 s.

#### Page No 230:

#### Question 6.5:

A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s^{−1 }about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

#### Answer:

Length of the rod, *l* = 1 m

Angular frequency,*ω* = 400 rad/s

Magnetic field strength, *B* = 0.5 T

One end of the rod has zero linear velocity, while the other end has a linear velocity of *l**ω**.*

Average linear velocity of the rod,

Emf developed between the centre and the ring,

Hence, the emf developed between the centre and the ring is 100 V.

#### Page No 230:

#### Question 6.6:

A
circular coil of radius 8.0 cm and 20 turns is rotated about its
vertical diameter with an angular speed of 50 rad s^{−1}
in a uniform horizontal magnetic field of magnitude 3.0×10^{−2}
T. Obtain the maximum and average emf induced in the coil. If the
coil forms a closed loop of resistance 10Ω,
calculate the maximum value of current in the coil. Calculate the
average power loss due to Joule heating. Where does this power come
from?

#### Answer:

Max induced *emf *=
0.603 V

Average induced *emf*
= 0 V

Max current in the coil = 0.0603 A

Average power loss = 0.018 W

(Power comes from the external rotor)

Radius of the circular
coil, *r* = 8 cm = 0.08 m

Area of the coil, *A*
= π*r*^{2} = π
× (0.08)^{2 }m^{2}

Number of turns on the
coil, *N* = 20

Angular speed, *ω*
= 50 rad/s

Magnetic field
strength, *B* = 3 ×
10^{−2} T

Resistance of the loop,
*R* = 10 Ω

Maximum induced *emf*
is given as:

*e* = *N**ω**
AB*

= 20 ×
50 × π
× (0.08)^{2} ×
3 × 10^{−2}

= 0.603 V

The maximum *emf*
induced in the coil is 0.603 V.

Over a full cycle, the
average *emf* induced in the coil is zero.

Maximum current is given as:

Average power loss due to joule heating:

The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent. It must supply a torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.

#### Page No 230:

#### Question 6.7:

A
horizontal straight wire 10 m long extending from east to west is
falling with a speed of 5.0 m s^{−1},
at right angles to the horizontal component of the earth’s
magnetic field, 0.30 × 10^{−4}
Wb m^{−2}.

**(a)
** What is the instantaneous value of
the emf induced in the wire?

**(b)
** What is the direction of the emf?

**(c)
** Which end of the wire is at the
higher electrical potential?

#### Answer:

Length
of the wire, *l*
= 10 m

Falling
speed of the wire, *v*
= 5.0 m/s

Magnetic
field strength, *B*
= 0.3 ×
10^{−4}
Wb m^{−2}

**(a) **Emf
induced in the wire,

*e*
= *Blv*

**(b) **Using
Fleming’s right hand rule, it can be inferred that the
direction of the induced emf is from West to East.

**(c) **The
eastern end of the wire is at a higher potential.

#### Page No 230:

#### Question 6.8:

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.

#### Answer:

Initial
current, *I*_{1}
= 5.0 A

Final
current, *I*_{2}
= 0.0 A

Change in current,

Time
taken for the change, *t*
= 0.1 s

Average
emf, *e*
= 200 V

For
self-inductance (*L)*
of the coil, we have the relation for average emf as:

*e*
= *L*

Hence, the self induction of the coil is 4 H.

#### Page No 230:

#### Question 6.9:

A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?

#### Answer:

Mutual
inductance of a pair of coils, *µ*
= 1.5 H

Initial
current, *I*_{1}
= 0 A

Final
current *I*_{2}
= 20 A

Change in current,

Time taken for the
change, *t* = 0.5 s

Induced emf,

Where is the change in the flux linkage with the coil.

Emf is related with mutual inductance as:

Equating equations (1) and (2), we get

Hence, the change in the flux linkage is 30 Wb.

#### Page No 230:

#### Question 6.10:

A
jet plane is travelling towards west at a speed of 1800 km/h. What is
the voltage difference developed between the ends of the wing having
a span of 25 m, if the Earth’s magnetic field at the location
has a magnitude of 5 × 10^{−4}
T and the dip angle is 30°.

#### Answer:

Speed of the jet plane,
*v* = 1800 km/h = 500 m/s

Wing spanof jet
plane,* l *= 25 m

Earth’s magnetic
field strength, *B* = 5.0 ×
10^{−4} T

Angle of dip,

Vertical component of Earth’s magnetic field,

*B*_{V}
= *B* sin

= 5 ×
10^{−4} sin 30°

= 2.5 ×
10^{−4} T

Voltage difference between the ends of the wing can be calculated as:

*e* = (*B*_{V})
× *l × v*

= 2.5 ×
10^{−4} × 25 ×
500

= 3.125 V

Hence, the voltage difference developed between the ends of the wings is

3.125 V.

#### Page No 231:

#### Question 6.11:

Suppose the loop in
Exercise 6.4 is stationary but the current feeding the electromagnet
that produces the magnetic field is gradually reduced so that the
field decreases from its initial value of 0.3 T at the rate of 0.02 T
s^{−1}. If the cut is joined and the loop has a
resistance of 1.6 Ω how
much power is dissipated by the loop as heat? What is the source of
this power?

#### Answer:

Sides of the rectangular loop are 8 cm and 2 cm.

Hence, area of the rectangular wire loop,

*A* = length ×
width

= 8 ×
2 = 16 cm^{2}

= 16 ×
10^{−4} m^{2}

Initial value of the magnetic field,

Rate of decrease of the magnetic field,

*Emf* developed in
the loop is given as:

Where,

= Change in flux through the loop area

= *AB*

Resistance of the loop,
*R* = 1.6 Ω

The current induced in the loop is given as:

Power dissipated in the loop in the form of heat is given as:

The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.

#### Page No 231:

#### Question 6.12:

A
square loop of side 12 cm with its sides parallel to X and Y axes is
moved with a velocity of 8 cm s^{−1}
in the positive *x-*direction
in an environment containing a magnetic field in the positive
*z*-direction.
The field is neither uniform in space nor constant in time. It has a
gradient of 10^{−3}
T cm^{−1}
along the negative *x-*direction
(that is it increases by 10^{−
3} T cm^{−1}
as one moves in the negative *x*-direction),
and it is decreasing in time at the rate of 10^{−3}
T s^{−1}.
Determine the direction and magnitude of the induced current in the
loop if its resistance is 4.50 mΩ.

#### Answer:

Side of the square
loop, *s* = 12 cm = 0.12 m

Area of the square
loop, *A* = 0.12 × 0.12
= 0.0144 m^{2}

Velocity of the loop, *v*
= 8 cm/s = 0.08 m/s

Gradient of the
magnetic field along negative *x*-direction,

And, rate of decrease of the magnetic field,

Resistance of the loop,

Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:

Rate of change of the
flux due to explicit time variation in field *B *is given as:

Since the rate of
change of the flux is the induced *emf*, the total induced *emf
*in the loop can be calculated as:

∴Induced current,

Hence, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.

#### Page No 231:

#### Question 6.13:

It
is desired to measure the magnitude of field between the poles of a
powerful loud speaker magnet. A small flat search coil of area 2 cm^{2}
with 25 closely wound turns, is positioned normal to the field
direction, and then quickly snatched out of the field region.
Equivalently, one can give it a quick 90° turn to bring its plane
parallel to the field direction). The total charge flown in the coil
(measured by a ballistic galvanometer connected to coil) is 7.5 mC.
The combined resistance of the coil and the galvanometer is 0.50 Ω.
Estimate the field strength of magnet.

#### Answer:

Area of the small
flat search coil, *A* = 2 cm^{2 }= 2 ×
10^{−4} m^{2}

Number of turns on the
coil, *N* = 25

Total charge flowing in
the coil, *Q* = 7.5 mC = 7.5 ×
10^{−3} C

Total resistance of the
coil and galvanometer, *R* = 0.50 Ω

Induced current in the coil,

Induced emf is given as:

Where,

= Charge in flux

Combining equations (1) and (2), we get

Initial flux through
the coil,
= *BA*

Where,

*B* = Magnetic
field strength

Final flux through the coil,

Integrating equation (3) on both sides, we have

But total charge,

Hence, the field strength of the magnet is 0.75 T.

#### Page No 231:

#### Question 6.14:

Figure
6.20 shows a metal rod PQ resting on the smooth rails AB and
positioned between the poles of a permanent magnet. The rails, the
rod, and the magnetic field are in three mutual perpendicular
directions. A galvanometer G connects the rails through a switch K.
Length of the rod = 15 cm, *B *=
0.50 T, resistance of the closed loop containing the rod = 9.0
mΩ. Assume
the field to be uniform.

**(a) ** Suppose
K is open and the rod is moved with a speed of 12 cm s^{−1}
in the direction shown. Give the polarity and magnitude of the
induced emf.

**(b) ** Is
there an excess charge built up at the ends of the rods when

K is open? What if K is closed?

**(c) ** With K
open and the rod moving uniformly, there is *no
net force *on the electrons in the
rod PQ even though they do experience magnetic force due to the
motion of the rod. Explain.

**(d) ** What
is the retarding force on the rod when K is closed?

**(e) ** How
much power is required (by an external agent) to keep the rod moving
at the same speed (=12 cm s^{−1})
when K is closed? How much power is required when K is open?

**(f) ** How
much power is dissipated as heat in the closed circuit?

What is the source of this power?

**(g) ** What
is the induced emf in the moving rod if the magnetic field is
parallel to the rails instead of being perpendicular?

#### Answer:

Length of the rod,* l*
= 15 cm = 0.15 m

Magnetic field
strength,* B* = 0.50 T

Resistance of the
closed loop, *R* = 9 mΩ
= 9 × 10^{−3} Ω

**(a)** Induced emf = 9 mV; polarity of the induced emf is such
that end *P* shows positive while end *Q* shows negative
ends.

Speed
of the rod, *v* = 12 cm/s = 0.12 m/s

Induced emf is given as:

*e*
= *Bvl*

*=
*0.5 × 0.12 ×
0.15

=
9 × 10^{−3} v

= 9 mV

The
polarity of the induced emf is such that end *P* shows positive
while end *Q* shows negative ends.

**(b) **Yes; when key K is closed, excess charge is maintained by
the continuous flow of current.

When key K is open, there is excess charge built up at both ends of the rods.

When key K is closed, excess charge is maintained by the continuous flow of current.

**(c) **Magnetic force is cancelled by the electric force set-up
due to the excess charge of opposite nature at both ends of the rod.

There is no net force on the electrons in rod PQ when key K is open and the rod is moving uniformly. This is because magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rods.

**(d) **Retarding
force exerted on the rod, *F* = *IBl*

Where,

*I*
= Current flowing through the rod

**(e)** 9 mW; no
power is expended when key K is open.

Speed
of the rod, *v* = 12 cm/s = 0.12 m/s

Hence, power is given as:

When key K is open, no power is expended.

**(f)** 9 mW; power
is provided by an external agent.

Power
dissipated as heat = *I*^{2} *R*

= (1)^{2} × 9 ×
10^{−3}

= 9 mW

The source of this power is an external agent.

**(g)** Zero

In this case, no emf is induced in the coil because the motion of the rod does not cut across the field lines.

#### Page No 232:

#### Question 6.15:

An
air-cored solenoid with length 30 cm, area of cross-section 25 cm^{2}
and number of turns 500, carries a current of 2.5 A. The current is
suddenly switched off in a brief time of 10^{−3}
s. How much is the average back emf induced across the ends of the
open switch in the circuit? Ignore the variation in magnetic field
near the ends of the solenoid.

#### Answer:

Length of the solenoid,
*l* = 30 cm = 0.3 m

Area of cross-section,
*A* = 25 cm^{2} = 25 ×
10^{−4 }m^{2}

Number of turns on the
solenoid, *N* = 500

Current in the
solenoid, *I* = 2.5 A

Current flows for time,
*t* = 10^{−3} s

Average back emf,

Where,

= Change in flux

= *NAB* …
(2)

Where,

*B* = Magnetic
field strength

Where,

= Permeability of free space = 4π
× 10^{−7} T m
A^{−1}

Using equations (2) and (3) in equation (1), we get

Hence, the average back emf induced in the solenoid is 6.5 V.

#### Page No 232:

#### Question 6.16:

**(a)** Obtain
an expression for the mutual inductance between a long straight wire
and a square loop of side *a *as
shown in Fig. 6.21.

**(b)** Now
assume that the straight wire carries a current of 50 A and the loop
is moved to the right with a constant velocity, *v
*= 10 m/s.

Calculate
the induced emf in the loop at the instant when *x
*= 0.2 m.

Take
*a *=
0.1 m and assume that the loop has a large resistance.

#### Answer:

**(a) **Take a small element *dy* in the loop at a distance *y*
from the long straight wire (as shown in the given figure).

Magnetic flux associated with element

Where,

*dA*
= Area of element *dy *= *a dy*

*B*
= Magnetic field at distance *y *

*I*
= Current in the wire

= Permeability of free space = 4π
× 10^{−7} T m
A^{−1}

*y*
tends from *x* to
.

**(b) **Emf induced
in the loop, *e* = *B’av*

Given,

*I
*= 50 A

*x*
= 0.2 m

*a*
= 0.1 m

*v*
= 10 m/s

#### Page No 232:

#### Question 6.17:

A line charge λ per
unit length is lodged uniformly onto the rim of a wheel of mass *M
*and radius *R*. The wheel has light non-conducting spokes
and is free to rotate without friction about its axis (Fig. 6.22). A
uniform magnetic field extends over a circular region within the rim.
It is given by,

**B **= − B_{0} **k **(*r *≤ *a*;
*a *< *R*)

= 0 (otherwise)

What is the angular velocity of the wheel after the field is suddenly switched off?

#### Answer:

Line charge per unit length

Where,

*r *= Distance of
the point within the wheel

Mass of the wheel = *M*

Radius of the wheel = *R*

Magnetic field,

At distance *r*,themagnetic force is balanced by the centripetal force
i.e.,

∴Angular velocity,

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