An electron is moving in a circular orbit of radius R with an angular acceleration \alpha At the center of the orbit is kept a conducting loop of radius r , (r< < R) The e.m.f induced in the smaller loop due to the motion of the electron is:

Using the formulas- I=\dfrac { e }{ T } =\dfrac { e\omega }{ 2\pi } \\ B\quad at\quad center=\dfrac { { \mu }_{ 0 }I }{ 2R } =\dfrac { \mu _{ 0 }e\omega }{ 4\pi R } \\ \phi =B.\pi { r }^{ 2 }=\dfrac { { \mu }_{ 0 } }{ 4\pi R } e\omega { \pi r }^{ 2 }\\ \\ =\dfrac { { \mu }_{ 0 }{ e\omega }{ r }^{ 2 } }{ 4R } \\ e=-\dfrac { d\phi }{ dt } =-\dfrac { \mu e{ r }^{ 2 }\alpha }{ 4R } \quad \quad (Ans)\\ \\ \\

Electromagnetic Induction - Easy Questions

Q: A circular coil of n turns is kept in a uniform magnetic field such that the plane of the coil is perpendicular to the field. The magnetic flux associated with the coil is now \phi . Now the coil is opened and made into another circular coil of twice the radius of the previous coil and kept in the same field such that the plane of the coil is perpendicular to the field. The magnetic flux associated with this coil now is:

flux(\phi )=nBA =nB\pi { r }^{ 2 } Let total length of coil is l l=n(2\pi r) Now, since same wire length has double radius 80, l={ n }_{ 1 }(2\pi 2r) l={ n }_{ 1 }=\dfrac { n }{ 2 } Now, flux =\dfrac { n }{ 2 } BA =\dfrac { n }{ 2 } B\pi4{ r }^{ 2 } = 2nB\pi { r }^{ 2 } = 2\phi So, new flux = 2\phi

Q: An electron is moving in a circular orbit of radius R with an angular acceleration \alpha At the center of the orbit is kept a conducting loop of radius r , (r< < R) The e.m.f induced in the smaller loop due to the motion of the electron is:

Using the formulas- I=\dfrac { e }{ T } =\dfrac { e\omega }{ 2\pi } \\ B\quad at\quad center=\dfrac { { \mu }_{ 0 }I }{ 2R } =\dfrac { \mu _{ 0 }e\omega }{ 4\pi R } \\ \phi =B.\pi { r }^{ 2 }=\dfrac { { \mu }_{ 0 } }{ 4\pi R } e\omega { \pi r }^{ 2 }\\ \\ =\dfrac { { \mu }_{ 0 }{ e\omega }{ r }^{ 2 } }{ 4R } \\ e=-\dfrac { d\phi }{ dt } =-\dfrac { \mu e{ r }^{ 2 }\alpha }{ 4R } \quad \quad (Ans)\\ \\ \\

Q: A conductor ABOCD moves along its bisector with a velocity of 1 ms ^{-1} through a perpendicular magnetic field of 1 Wbm ^{-2} , as shown in the figure. If all the four sides are of 1 m length each, then the induced emf between point A and D is :

Given: B = 1 Wbm^{-2} ; v = 1 m/sFrom figure, x = 1\times sin45^o = \dfrac{1}{\sqrt{2}} Thus, l = 2x = \sqrt{2} Induced emf between A and D: \mathcal{E} = Bvl \therefore \mathcal{E} = 1\times 1\times \sqrt{2} = 1.41\ V

Q: A square frame of side 10 cm and a long straight wire carrying current 1 A are in the plane of the paper. Starting from close to the wire, the frame moves towards the right with a constant speed of 10 ms^{-1} (see figure). The e.m.f induced in the frame at the time the left arm is at x=10cm from the wire is :

In given fig x=15 because left arm of the frame is at 10cm from the wire. and a=10cm .emf in AD \Longrightarrow e_1 \Longrightarrow \dfrac{a\mu _oiv}{2\pi (x-a/2)} emf in EF \Longrightarrow e_2 \Longrightarrow \dfrac{a\mu _oiv}{2\pi (x+a/2)} Net emf =e_1 - e_2 = \dfrac{a\mu _oiv}{2\pi (x-a/2)} - \dfrac{a\mu _oiv}{2\pi (x+a/2)} putting the given values into the equation we get,emf induced =1.3\mu V\approx 1\mu V

Q: What amount of power is generated in that resistance?

The liquid is conducting and hence contains free electrons. The electrons in the field \vec{B} experiences force Bev . This force is in y-direction and the electrons drift in this direction and this builds up a potential between the plates in such a way that \dfrac{V}{d} = Bev The resistance of the liquid between the plates is \dfrac{\rho d }{S} The total resistance is \dfrac{\rho d }{S} + R The current in the external resistance is I = \dfrac{V}{ R + \dfrac{\rho d }{S} }=\dfrac{Bvd}{R + \dfrac{\rho d }{S}} Power P = I^2R = \dfrac{V}{ R + \dfrac{\rho d }{S} }=(\dfrac{Bvd}{R + \dfrac{\rho d }{S}})^2R

Q: The force required to keep the rod moving with the speed of 3 ms ^{-1} (along + x-axis) is :(Take K = 4.5 N)

E=4.5V R=2.5+0.5=3\Omega i=\frac{4.5}{3}=1.5A Magnetic force m wire =i\overrightarrow {l}\times \overrightarrow{B} =(1.5)(1.5)(2)(-)(\frac{\sqrt{3}i}{2}+\frac{j}{2})(\widehat{k}) =(-)4.5\left ( \frac{-\sqrt{3}\widehat{j}}{2}+\frac{\widehat{j}}{2} \right ) The force required to keep it in motion must balance this magnetic force. \therefore \overrightarrow{F}=(-)(-)4.5\left ( \frac{\widehat{i}}{2}-\frac{\sqrt{3}\widehat{j}}{2} \right ) =4.5\left ( \frac{\widehat{i}}{2}-\frac{\sqrt{3}\widehat{j}}{2} \right )

Q: A long solenoid of length L , cross section A having N_1 turns has wound about its centre with a small coil of N_2 turns as shown in Fig. The mutual inductance of two circuits is:

\phi_2=N_2B_1A \phi_2=N_2\times \dfrac {\mu_0N_1I_1}{L}\times A \phi_2=\dfrac {\mu_0N_1N_2A}{L}I_1 Comparing with: \phi_2=MI_1 We get: M_2=\dfrac {\mu_0N_1N_2A}{L}

Q: The approximate formula expressing the formula of mutual inductance of two thin coaxial loops of the same radius a when their centers are separated by a distance l with l > > a is

Let I = current in one loop. The magnetic field at the centre of the other co-axial loop at a distance l from the centre of the first loop is B=\dfrac {\mu_0}{4\pi}\dfrac {2I\pi a^2}{(a^2+l^2)^{3/2}} where, p_m=I\pi a^2 = magnetic moment of the loopFlux through the other loop is \phi_{12}=B\pi a^2 =\dfrac {\mu_0}{4\pi}\dfrac {2\pi a^2I}{(a^2+l^2)^ {3/2}}\pi a^2 or M=\dfrac {\phi_{12}}{I}=\dfrac {\mu_0}{4\pi}\dfrac {2\pi^2a^4}{(a^2+l^2)^{3/2}} =\dfrac {\mu_0\pi a^4}{2(a^2+l^2)^{3/2}}=\dfrac {\mu_0\pi a^4}{2l^3} (a < < l)

Q: Consider the circuit shown in figure. (a) Find the current through the battery a long time after the switch S is closed. (b) Suppose the switch is again opened at t=0 . What is the time constant of the discharging circuit?

(a) The inductor does not work in DC. When the switch is closed the current charges so at first inductor works. But after a long time the current flowing is constantThus effect of inductance vanishes \quad i=\cfrac { E }{ { R }_{ net } } =\cfrac { E }{ \cfrac { { R }_{ 1 }{ R }_{ 2 } }{ { R }_{ 1 }+{ R }_{ 2 } } } =\cfrac { E\left( { R }_{ 1 }+{ R }_{ 2 } \right) }{ { R }_{ 1 }{ R }_{ 2 } } (b) When the switch is opened the resistors are in series \tau =\cfrac { L }{ { R }_{ net } } =\cfrac { L }{ { R }_{ 1 }+{ R }_{ 2 } }

Q: An inductance coil is connected to an ac source through a 60\Omega ,resistance in series. The source voltage, voltage across the coil and voltage across the resistance are found to be 33V, 27V abd 12V respectively. Therefore, the resistance of the coil is

I=\dfrac{12}{60}=\dfrac{1}{5} V^2_S=V^2_L+(V_R+V_{60})^2 V^2_S=V^2_L+V^2_R+2V_R V_{60}+V^2_{60} (33)^2=(27)^2+2V_R(12)+(12)^2 1089=729+24V_R+144 24V_R=216 V_R=\dfrac{216}{24}=9 IR = 9 R=\dfrac{9}{I}=\dfrac{1}{\frac{1}{5}}=45\Omega

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Easy Questions

11 min read

Electromagnetic Induction

- It is always better to start confident. Let's practice easy questions to serve the purpose

Problem on Magnetic Flux.

A circular coil of n turns is kept in a uniform magnetic field such that the plane of the coil is perpendicular to the field. The magnetic flux associated with the coil is now

ϕ

. Now the coil is opened and made into another circular coil of twice the radius of the previous coil and kept in the same field such that the plane of the coil is perpendicular to the field. The magnetic flux associated with this coil now is:

ϕ

ϕ

\frac{ϕ}{4}

\frac{ϕ}{2}

View solution

Induced emf in a closed loop.

An electron is moving in a circular orbit of radius R with an angular acceleration

α

At the center of the orbit is kept a conducting loop of radius

r

(r < < R)

The e.m.f induced in the smaller loop due to the motion of the electron is:

zero, since charge on electron in constant

\frac{μ _{0} e r ^{2}}{4 R} α

\frac{μ _{0} e r ^{2}}{4 π R} α

none of these

View solution

A conductor ABOCD moves along its bisector with a velocity of 1 ms

^{- 1}

through a perpendicular magnetic field of 1 Wbm

^{- 2}

, as shown in the figure. If all the four sides are of 1 m length each, then the induced emf between point A and D is :

zero

1.41 V

0.71 V

none of the above

View solution

Induced emf in a coil caused by varying flux due to magnetic field.

A square frame of side

10 c m

and a long straight wire carrying current

1 A

are in the plane of the paper. Starting from close to the wire, the frame moves towards the right with a constant speed of

10 m s^{- 1}

(see figure). The e.m.f induced in the frame at the time the left arm is at

x = 10 c m

from the wire is :

2 μ V

1 μ V

0.75 μ V

0.5 μ V

View solution

Problems On Motional EMF.

What amount of power is generated in that resistance?

P = \frac{2 v ^{2} B ^{2} d ^{2} R}{( R + \frac{2 ρ d}{S} ) ^{2}}

P = \frac{2 v ^{2} B ^{2} d ^{2} R}{( R + \frac{ρ d}{S} ) ^{2}}

P = \frac{v ^{2} B ^{2} d ^{2} R}{( R + \frac{2 ρ d}{S} ) ^{2}}

P = \frac{v ^{2} B ^{2} d ^{2} R}{( R + \frac{ρ d}{S} ) ^{2}}

View solution

The force required to keep the rod moving with the speed of 3 ms

^{- 1}

(along + x-axis) is :(Take K = 4.5 N)

(\frac{1}{2} \hat{i} + \frac{3}{2} \hat{j})

(\frac{3}{2} \hat{i} - \frac{1}{2} \hat{j})

(\frac{3}{2} \hat{i} + \frac{1}{2} \hat{j})

(\frac{1}{2} \hat{i} - \frac{3}{2} \hat{j})

View solution

Problem on Mutual Inductance.

A long solenoid of length

L

, cross section

A

having

N_{1}

turns has wound about its centre with a small coil of

N_{2}

turns as shown in Fig. The mutual inductance of two circuits is:

\frac{μ _{0} A ( N _{1} / N _{2} )}{L}

\frac{μ _{0} A ( N _{1} N _{2} )}{L}

μ_{0} A N_{1} N_{2} L

\frac{μ _{0} A N _{1}^{2} N _{2}}{L}

View solution

Calculate mutual inductance of two inductors of different shapes.

The approximate formula expressing the formula of mutual inductance of two thin coaxial loops of the same radius

a

when their centers are separated by a distance

l

with

l > > a

\frac{1}{2} \frac{μ _{0} π a ^{4}}{l ^{3}}

\frac{1}{2} \frac{μ _{0} a ^{4}}{l ^{2}}

\frac{μ _{0}}{4 π} \frac{π a ^{2}}{l ^{2}}

\frac{μ _{0}}{π} \frac{a ^{4}}{l ^{3}}

View solution

Problem on combination of Inductor and Kirchhoff's Law

Consider the circuit shown in figure. (a) Find the current through the battery a long time after the switch

S

is closed. (b) Suppose the switch is again opened at

t = 0

. What is the time constant of the discharging circuit?

View solution

Problem on Inductors

An inductance coil is connected to an

a c

source through a

60 Ω

,resistance in series. The source voltage, voltage across the coil and voltage across the resistance are found to be

33 V, 27 V

abd

12 V

respectively. Therefore, the resistance of the coil is

30 Ω

45 Ω

105 Ω

75 Ω

View solution

Basic problem on dc and ac motors.

An electric motor connected to a

120 V

60.0 H z

ac outlet does mechanical work at the rate of

0.100 h p

(

1 h p = 746 W

). (a) If the motor draws an

r m s

current of

0.650 A

, what is its effective resistance, relative to power transfer? (b) Is this the same as the resistance of the motors coils, as measured with an ohmmeter with the motor disconnected from the outlet?

View solution

Problem on transformer (EMF and current in AC generator).

Stepping up of voltage by a factor of 100,reduce the current by a factor of

100

leaves the current unchanged

increase the current by a factor of 100

View solution

Problem on DC Generator.

A neon lamp is connected to a voltage a.c. source. The voltage is gradually increased from zero volt. It is observed that the neon flashes at 50 V. The a.c. source is now replaced by a variable dc source and the experiment is repeated. The neon bulb will flash at

50V

70V

100V

35V

View solution