Q: The magnitude of the induced electric field inside the solenoid, at a distance r < R from its long central axis is :

Magnetic field inside the solenoid , B=\mu_0nI Flux through the area of radius r, \phi =\int { B.ds } =\mu_0nI\left( \pi { r }^{ 2 } \right) \left| \oint { \overrightarrow { E. } dl } \right| =\left| \dfrac { d\phi m }{ dt } \right| \overrightarrow { E } .\left( 2\pi r \right) =\left| \mu _0n\left( \pi { r }^{ 2 } \right) \dfrac { dI }{ dt } \right| \overrightarrow { E } =\left| \dfrac { \mu_0nr }{ 2 } \dfrac { d\left( I_{max}\cos { \omega t } \right) }{ dt } \right| \overrightarrow { E } =\dfrac { \mu_0nI_{max} \omega }{ 2 } r\sin { \omega t }

Q: A square wire frame of side a is placed a distance b away from a long straight conductor carrying current I . The frame has resistance R and self inductance L . The frame is rotated by 180^o about OO' as shown in figure. Find the electric charge flown through the frame.

\displaystyle i = \dfrac{1}{R} \left [ \dfrac{d \psi}{dt} + L \dfrac{di}{dt} \right ] \displaystyle q = \int idt = \dfrac{1}{R} [\Delta \psi + 0] = \dfrac{\Delta \psi}{R} = \dfrac{1}{R} \int_{b - a}^{b + a} Ba.dx \displaystyle = \dfrac{1}{R} \int_{b-a}^{b +a } \dfrac{\mu_0 ia}{2 \pi x} dx = \dfrac{\mu_0 ia}{2 \pi R} \log_e \dfrac{b + a}{b -a}

Q: A non-conducting ring having charge q uniformly distributed over its circumference is placed on a rough horizontal surface. A vertical time varying magnetic field B = 4t^{2} is switched on at time t=0 . Mass of the ring is 'm' and radius is R. The ring starts rotating after 2 s. The coefficient of friction between the ring and the table is:

e=\dfrac{-d\phi}{dt}=-\dfrac{d(B.A)}{dt}=\pi R^2\dfrac{d(4t^2)}{dt}=-8\pi R^2t min. force required to move ring is =\mu mg where \mu is coeff. of frictionat t=2 Sec emf,\ V=16\pi R^2 Force on ring = qE=\dfrac { qV }{ 2\pi R } =\dfrac { 16q\pi { R }^{ 2 } }{ 2\pi R } =8qR=\mu mg So, \mu=\dfrac{8qR}{mg}

Q: A uniform but time-varying magnetic field B(t) exists in a circular region of radius a and is directed into the plane of the paper as shown in the figure. The magnitude of the induced electric field at point P at a distance r form the centre of the circular region

\int E\cdot dl=-\dfrac{d\Phi}{dt} E\cdot 2\pi r=-\pi a^2\dfrac{dB}{dt} E=-\dfrac{1}{r}(constant) E \ \alpha \ \dfrac{1}{r} (and -ve sign indicates that E decreases)

Q: Figure shows a square loop of side 0.5m and resistance 10\Omega . The magnetic field has a magnitude B=1.0T . The work done in pulling the loop out of the field slowly and uniformly in 2.0s is

emf = B l \dfrac{l}{2} work done in pulling out = \dfrac{emf^2}{R} \times t =\dfrac{B^2 l^2 \dfrac{l^2}{2^2} }{R} \times 2= 3.125 mJ

Q: A metallic ring of mass 2 kg and radius 1m with a uniform metallic spoke of same mass 2kg and length 1 m is rotated about its axis with angular velocity 1 rev/sec . in a perpendicular uniform magnetic field B of magnitude 10T as shown in figure. The central end of the spoke is connected to the rim of the wheel through a resistor R of magnitude \pi \Omega as shown. The resistor does not rotate, its one end is always at the center of the ring and other end is always in contact with the ring. A force F as shown is needed to maintain constant angular velocity of the spoke then, F is equal to (The ring and the spoke has zero resistance)

F_b = BIL Induced current I=\dfrac{(B\omega r^2 /2}{R} \therefore F_b=B(\dfrac{B^2 \omega r^3}{2R})r=\dfrac{B^2 \omega r^3}{2R} To maintain constant angular velocity : F(r) =F_B(r/2) \Rightarrow F=\dfrac{F_B}{2}=\dfrac{B^2\omega r^3}{4R}=\dfrac{(10)^2 \times 2 \pi \times (1)^3}{4 \times \pi}=50 N

Q: A square loop of side a is rotating about its diagonal with angular velocity \omega in a perpendicular magnetic field \vec B . It has 10 turns. The emf induced is

\phi=nBA cos\theta =10 Ba^2 cos\omega t e=-\dfrac {d\phi}{dt} =-\dfrac {d}{dt}(10 Ba^2 cos \omega t) =10 Ba^2 sin\omega t(\omega)

Q: A circular wire loop of radius R is placed in the x-y plane centered at the origin O. A square loop of side a (a << R) having two turns is placed with its center at z = \sqrt{3}R along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of 45^{\circ} with respect to the z-axis. lf the mutual inductance between the loops is given by \dfrac{\mu_{0}a^2}{2^{p/2}R} , then the value of p is

\displaystyle M= \frac{N \phi}{I} \displaystyle = \dfrac {2 \dfrac{\mu_{o}IR^{2}}{2(8R^{3})}a^{2} \cos45^o}{I} \displaystyle = \dfrac{\mu_{o}a^{2}}{8 R\times 2^{1/2}}= \frac{\mu_{o}a^{2}}{R\times 2^{7/2}} So, p= 7

Q: In the circuit shown (fig), the coil has inductance and resistance. When X is joined to Y, the time constant is \tau during the growth of current. When the steady state is reached, heat is produced in the coil at a rate P. X is now joined to Z. After joining X and Z :

In steady state: P= I^2 R Energy stored in inductor =\dfrac{1}{2} L I^2 \tau = L/R After connecting x and z, the whole energy of inductor will go into heat so heat produced H=\dfrac{1}{2}LI^2=\dfrac{1}{2}L\dfrac{P}{R}=\dfrac{1}{2}P\tau

Question 1

The magnitude of the induced electric field inside the solenoid, at a distance  r < R  from its long central axis is :

Accepted Answer

Magnetic field inside the solenoid , B=\mu_0nI Flux through the area of radius r,   \phi =\int { B.ds } =\mu_0nI\left( \pi { r }^{ 2 } \right)     \left| \oint { \overrightarrow { E. } dl }  \right| =\left| \dfrac { d\phi m }{ dt }  \right|   \overrightarrow { E } .\left( 2\pi r \right) =\left| \mu _0n\left( \pi { r }^{ 2 } \right) \dfrac { dI }{ dt }  \right|    \overrightarrow { E } =\left| \dfrac { \mu_0nr }{ 2 } \dfrac { d\left( I_{max}\cos { \omega t }  \right)  }{ dt }  \right|    \overrightarrow { E } =\dfrac { \mu_0nI_{max} \omega  }{ 2 } r\sin { \omega t }

Question 2

A square wire frame of side  a  is placed a distance  b  away from a long straight conductor carrying current  I . The frame has resistance  R  and self inductance  L . The frame is rotated by  180^o  about OO' as shown in figure. Find the electric charge flown through the frame.

Accepted Answer

\displaystyle i = \dfrac{1}{R} \left [ \dfrac{d \psi}{dt} + L \dfrac{di}{dt} \right ] \displaystyle q = \int idt = \dfrac{1}{R} [\Delta \psi + 0] = \dfrac{\Delta \psi}{R} = \dfrac{1}{R} \int_{b - a}^{b + a} Ba.dx \displaystyle = \dfrac{1}{R} \int_{b-a}^{b +a } \dfrac{\mu_0 ia}{2 \pi x} dx = \dfrac{\mu_0 ia}{2 \pi R} \log_e \dfrac{b + a}{b -a}

Question 3

A non-conducting ring having charge q uniformly distributed over its circumference is placed on a rough horizontal surface. A vertical time varying magnetic field

B = 4 t^{2}

is switched on at time

t = 0

. Mass of the ring is 'm' and radius is R. The ring starts rotating after 2 s. The coefficient of friction between the ring and the table is:

Accepted Answer

e=\dfrac{-d\phi}{dt}=-\dfrac{d(B.A)}{dt}=\pi R^2\dfrac{d(4t^2)}{dt}=-8\pi R^2t min. force required to move ring is  =\mu mg where  \mu   is coeff. of frictionat   t=2 Sec          emf,\ V=16\pi R^2  Force on ring =  qE=\dfrac { qV }{ 2\pi R } =\dfrac { 16q\pi { R }^{ 2 } }{ 2\pi R } =8qR=\mu mg So,  \mu=\dfrac{8qR}{mg}

Question 4

A uniform but time-varying magnetic field

B (t)

exists in a circular region of radius

a

and is directed into the plane of the paper as shown in the figure. The magnitude of the induced electric field at point

P

at a distance

r

form the centre of the circular region

A

is zero

B

decreases as

\frac{1}{r}

C

increases as

r

D

decreases as

\frac{1}{r ^{2}}

View solution

Accepted Answer

\int E\cdot dl=-\dfrac{d\Phi}{dt} E\cdot 2\pi r=-\pi a^2\dfrac{dB}{dt} E=-\dfrac{1}{r}(constant) E \ \alpha \ \dfrac{1}{r}       (and  -ve  sign indicates that  E  decreases)

Question 5

Figure shows a square loop of side  0.5m  and resistance  10\Omega . The magnetic field has a magnitude  B=1.0T . The work done in pulling the loop out of the field slowly and uniformly in  2.0s  is

Accepted Answer

emf =  B l \dfrac{l}{2}  work done in pulling out =   \dfrac{emf^2}{R} 	imes t   =\dfrac{B^2 l^2 \dfrac{l^2}{2^2} }{R} 	imes 2= 3.125 mJ

Question 6

A metallic ring of mass  2 kg  and radius  1m  with a uniform metallic spoke of same mass  2kg  and length  1 m  is rotated about its axis with angular velocity  1 rev/sec . in a perpendicular uniform magnetic field  B  of magnitude  10T  as shown in figure. The central end of the spoke is connected to the rim of the wheel through a resistor  R  of magnitude  \pi  \Omega  as shown. The resistor does not rotate, its one end is always at the center of the ring and other end is always in contact with the ring. A force  F  as shown is needed to maintain constant angular velocity of the spoke then,  F  is equal to (The ring and the spoke has zero resistance)

Accepted Answer

F_b = BIL  Induced current   I=\dfrac{(B\omega r^2 /2}{R}   	herefore F_b=B(\dfrac{B^2 \omega r^3}{2R})r=\dfrac{B^2 \omega r^3}{2R}  To maintain constant angular velocity :  F(r) =F_B(r/2)   \Rightarrow F=\dfrac{F_B}{2}=\dfrac{B^2\omega r^3}{4R}=\dfrac{(10)^2 	imes 2 \pi 	imes (1)^3}{4 	imes \pi}=50 N

Question 7

A square loop of side a is rotating about its diagonal with angular velocity  \omega  in a perpendicular magnetic field  \vec B . It has 10 turns. The emf induced is

Accepted Answer

\phi=nBA cos	heta =10 Ba^2 cos\omega t e=-\dfrac {d\phi}{dt} =-\dfrac {d}{dt}(10 Ba^2 cos \omega t) =10 Ba^2 sin\omega t(\omega)

Question 8

A circular loop of radius  r  is placed at the centre of current carrying conducting square loop of side  a . If both loops are coplanar and  a >> r , then the mutual inductance between the loops will be:

Accepted Answer

Both loops are coplanar. Magnetic field at the center of outer square current carrying loop is { B }_{ 1 }=\dfrac { 2\sqrt { 2 } { \mu  }_{ 0 }I }{ \pi a }  where  a = length of side of square loop and r = radius of the circular loop.Given  a>>r ,The magnetic field through entire inner coil is  { B }_{ 1 } Magnetic flux through inner coil,  { \phi  }_{ 21 }={ B }_{ 1 }{ A }_{ 2 }         = \dfrac { 2\sqrt { 2 } { \mu  }_{ 0 }I }{ \pi  } \dfrac { { r }^{ 2 } }{ a }  ------ (1)    Mutual induction, M=  \dfrac { \phi  }{ { I }_{ 1 } }  From (1), M=  \dfrac { 2\sqrt { 2 } { \mu  }_{ 0 } }{ \pi  } \dfrac { { r }^{ 2 } }{ a }  Hence,  M\alpha \dfrac { { r }^{ 2 } }{ a }

Question 9

A circular wire loop of radius  R  is placed in the x-y plane centered at the origin O. A square loop of side  a (a << R)  having two turns is placed with its center at  z = \sqrt{3}R  along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of  45^{\circ}  with respect to the z-axis. lf the mutual inductance between the loops is given by  \dfrac{\mu_{0}a^2}{2^{p/2}R} , then the value of  p  is

Accepted Answer

\displaystyle M= \frac{N \phi}{I}   \displaystyle = \dfrac {2 \dfrac{\mu_{o}IR^{2}}{2(8R^{3})}a^{2} \cos45^o}{I} \displaystyle = \dfrac{\mu_{o}a^{2}}{8 R	imes 2^{1/2}}= \frac{\mu_{o}a^{2}}{R	imes 2^{7/2}} So,   p= 7

Question 10

In the circuit shown (fig), the coil has inductance and resistance. When X is joined to Y, the time constant is  	au  during the growth of current. When the steady state is reached, heat is produced in the coil at a rate P. X is now joined to Z. After joining X and Z :

Accepted Answer

In steady state:   P= I^2 R  Energy stored in inductor   =\dfrac{1}{2} L I^2  	au = L/R   After connecting x and z, the whole energy of inductor will go into heat so heat produced    H=\dfrac{1}{2}LI^2=\dfrac{1}{2}L\dfrac{P}{R}=\dfrac{1}{2}P	au

Question 11

Study the diagram. As soon as the switch  S  is closed, the current through the cell is  I_1 . After a long time the current through the cell is found to be  I_2 , then :

Accepted Answer

When switch  S  first closed, current through inductor is  I_1=0 : because it cannot change instantaneously. This means that the inductor acts like an open circuit, so all the voltage is across the inductor. As current increases, magnetic field strengthens and the changing magnetic field creates a back emf which acts to oppose the current in the inductor. This back emf will not stop the current completely, but it will slow it down.Eventually, the current in the inductor reaches full strength (as governed by the resistor and the voltage by Ohm's Law). When this happens, the current is no longer changing, so the voltage across the inductor is zero. So,  I_2=\frac{E_0}{R}

Electromagnetic Induction

Use of line integral of electric field

Circuit rotating perpendicular to the plane of magnetic field

Induced emf due to varying magnetic field

Power required to move a conductor in a magnetic field

Force required to move a conductor in a magnetic field

Problem in which circuit is rotating in the plane of magnetic field

Problem based on mutual inductance of different shape

Energy stored in an inductor due to a magnetic field

Behaviour of RL circuits at the time of switching