Q: A 3.56H inductor is placed in series with a 12.8\Omega resistor. An emf of 3.24V is then suddenly applied across the RL combination. At 0.278s after the emf is applied what is the rate at which energy is being delivered by the battery?

{ \tau }_{ L }=\cfrac { L }{ R } =\cfrac { 3.56 }{ 12.8 } =0.278s { i }_{ 0 }=\cfrac { V }{ R } =\cfrac { 3.24 }{ 12.8 } =0.253A After one time constant ( t=0.278s={\tau}_{C} ) \quad i=\left( 1-\cfrac { 1 }{ e } \right) { i }_{ 0 }=0.63{ i }_{ 0 }=0.16A Power supplied by battery =Ei P=(3.24\times (0.16)=0.518W

Question 1

Question 2

A wire as a parabola  y = 4x^2  is located in a uniform magnetic field of inductance B perpendicular to the XY plane. At t=0 a connection starts translation wise from the parabola apex with constant acceleration  \alpha . The induced emf in the loop, thus formed, as a function of y is:

Accepted Answer

\overrightarrow { B } =B\uparrow   Velocity of rod after travelling  dy  distance  u .we know that, S=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }   y=0+\dfrac { 1 }{ 2 } \alpha { t }^{ 2 }   ( t=\sqrt{\dfrac{2y}{\alpha}} )  \dfrac { dy }{ dt } =\alpha t   V=\alpha t=\sqrt { 2\alpha y }   emf induced   \varepsilon =VB(2x) =\sqrt { 2\alpha y } 	imes 2xB    x=\dfrac { \sqrt { y }  }{ 2 }   \varepsilon =\sqrt { 2\alpha  } .By

Question 3

An annular circular brass disk of inner radius

r

and outer radius

R

is rotating about an axis passing through its center and perpendicular to its plane with a uniform angular velocity

ω

in a uniform magnetic field of induction

B

normal to the plane of the disk. The induced emf between the inner and the outer edge of the annular disk is :

Accepted Answer

d\left( emf \right) =-Bwxdx emf=-\int _{ r }^{ R }{ Bwxdx }  =-Bw \int_{r}^{R}\dfrac { { x }^{ 2 } }{ 2 } =Bw\left( \dfrac { { R }^{ 2 }-{ r }^{ 2 } }{ 2 } \right)  Emf between inner and outer edge will be =Bw\left( \dfrac { { R }^{ 2 }-{ r }^{ 2 } }{ 2 } \right)

Question 4

A wire shaped as a semicircle of radius a is rotating about an axis PQ with a constant angular velocity

ω = \frac{1}{L C}

, with the help of an external agent. A uniform magnetic field B exists in space and is directed into the plane of the figure. (circuit part remains at rest (left part is at rest) ) . Then,

Accepted Answer

Let at time t the angle between magnetic field and area vector(semicircle) be  	heta ,Then   	heta =\omega t Magnetic flux at time t is: \phi =\vec {B}.\vec {S}=\dfrac{\pi a^{2}B}{2}\cos \omega t Induced emf at time t is: \varepsilon =-\dfrac{d\phi }{dt}=\dfrac{\pi Ba^{2}\omega }{2}\sin \omega t \varepsilon_o=\dfrac{\pi Ba^{2} }{2\sqrt{LC}} As  \omega = \dfrac{1}{\sqrt {LC}} , the circuit is in resonance. \left | Z ight |=R   \Rightarrow i _{0}=\dfrac{\pi Ba^{2} }{2R\sqrt{LC}}  which is the peak current. i_{rms}=\dfrac{i_{0}}{\sqrt{2}}   \Rightarrow i _{rms}=\dfrac{\pi Ba^{2} }{2R\sqrt{ 2 LC}} V_{C}=\frac{1}{2}CV_{0}^{2}ightarrow max. energy ,   V_{0}ightarrow peak voltage V_{0}= i_{0}X_{c}=\dfrac{i_{0}}{C\omega }=\dfrac{i_{0}\sqrt{LC}}{C} V_{C}=\dfrac{1}{2}C	imes\dfrac{\pi ^{2}B^{2}a^{4}}{4R^{2}C^{2}}= \dfrac{\pi ^{2}B^{2}a^{4}}{8R^{2}C}  P_{ext}=P_{dissipated}=\varepsilon _{0}i_{0}=\dfrac{\pi Ba^{2} }{2\sqrt{LC}}	imes \dfrac{\pi Ba^{2} }{2R\sqrt{LC}} P_{ext}=\dfrac{\pi ^{2}B^{2}a^{4}}{4LCR}

Question 5

A conducting rod of length  l  is hinged at point O. It is free to rotate in a vertical plane. There exists a uniform magnetic field  \vec { B }  in horizontal direction. The rod is released from the position shown. The potential difference between the two ends of the rod is proportional to:

Accepted Answer

V=\dfrac{B}{2}R^2\omega where  \omega=\sqrt{(\dfrac{2}{I})(mg\dfrac{l}{2}\sin	heta)} \omega\alpha \sqrt{\dfrac{\sin 	heta}{l}}       I=kl^2  ( I  - moment of intertia) 	herefore V  \alpha \ l^2	imes \sqrt{\dfrac{\sin	heta}{l}} V  \alpha l^{\dfrac{3}{2}} and    V  \alpha \sqrt{\sin	heta}

Question 6

An infinitely long wire is placed near a square loop as shown in the figure. Choose the correct options.

Accepted Answer

Hint :  \bullet For an infinitely long wire, the magnetic field is given as   B=\frac {\mu_o I}{2\pi r}  \bullet Mutual inductance is given as  M=\frac {\phi}{I} (a)  	extbf{Step 1 : Magnetic field due to infinite current carrying wire at distance x} The magnetic field of the infinite wire is-  \mathrm{B}=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{i}}{\mathrm{x}} 	extbf{Step2:Magnetic flux through an element} Considering an elemental length of width dx at a diatance of x from the straight wire.Area of element dS=adxThe magnetic field at dx is, \mathrm{B}_{\mathrm{x}}=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{i}}{\mathrm{x}} The magnetic flux through the region  of width dx is, \Rightarrow \mathrm{d} \phi=\left(\mathrm{B}_{\mathrm{x}}ight) \mathrm{d} \mathrm{S} \Rightarrow d\phi=\left(\frac{\mu_{0}}{2 \pi} \frac{\mathrm{i}}{\mathrm{x}}ight)(\mathrm{adx}) 	extbf{Step3:Total Flux through square loop} Taking integration on RHS from a to 2a we have, \Rightarrow \phi=\int_{\mathrm{a}}^{2 \mathrm{a}} \mathrm{d} \phi \Rightarrow \phi=\frac{\mu_{0} \mathrm{ia}}{2 \pi} \ln 2 	extbf{Step4:The mutual inductance between the long wire and square loop}   M=\frac{\phi}{I} \Rightarrow M=\frac{\mu_{0} \mathrm{a}}{2 \pi} \ln 2  (c)  Wire produces &#10752; magnetic field over the loop. if the loop is brought closer to the wire &#10752; magnetic field passing through the loop increases. Hence induced current produces &#10754; magnetic field so, induced current is clockwise.Therefore the correct options are both 'A' & 'C' is ( A ) The mutual inductance between the two is  \frac{\mu_{0} a}{2 \pi} \ln (2) . ( C ) If a constant current is passed in the straight wire in upward direction and loop is brought close to the wire then induced current in the loop is clockwise

Question 7

What is the potential difference  {V}_{ab}  in Volts across the inductor  {R}_{1} ?

Accepted Answer

At  t=0  , the current through inductor is  0  .Hence, the potential across  R_1=E=120V If there would be current through inductor then, there would be emf produced in the inductor and in that condition we have to treat the circuit as two cells in parallel one  E  and other due to inductor .Answer- (120V)

Question 8

The conducting rod  ab , as shown in figure makes contact with metal rails  ca  and  db . The apparatus is in a uniform magnetic field of  0.800T , perpendicular to the plane of the figure. Compare the rate at which mechanical work done by the force  (Fv)  with the rate at which thermal energy is developed in the circuit ( {I}^{2}R ).(in  W )

Accepted Answer

F={F}_{m}=ilB=\cfrac{e}{r}\cdot lB=\cfrac{Bvl}{r}\cdot Bl=\cfrac{{ B }^{ 2 }{ l }^{ 2 }}{r}\cdot v =\cfrac { { \left( 0.8 ight)  }^{ 2 }{ \left( 0.5 ight)  }^{ 2 } }{ { \left( 1.5 ight)  }^{  } } 	imes 7.5=0.8N F\cdot v=0.8	imes 7.5=6W {i}^{2}R={ \left( \cfrac { {(Bvl) }^2}{ R }  ight)  }=\cfrac { { B }^{ 2 }{ l }^{ 2 } }{ R } \cdot { v }^{ 2 } =\cfrac { { \left( 0.8 ight)  }^{ 2 }{ \left( 0.5 ight)  }^{ 2 } }{ { \left( 1.5 ight)  }^{  } } 	imes { 7.5 }^{ 2 }=6W

Question 9

In the figure there are two identical conducting rods each of length  'a'  rotating with angular speed  \omega  in the directions shown. One end of each rod touches a conducting ring. Magnetic field  B  exists perpendicular to the plane of the rings. The rods, the conducting rings and the lead wires are resistanceless. Find the magnitude and direction of current in the resistance  R .

Accepted Answer

emf induced in each loop is  \left| e \right|  e=\dfrac{1}{2} \omega Ba Total emf in the circuit is, 2e=\omega Ba Current  i=\dfrac{2e}{R}=\dfrac{\omega Ba}{R}  direction shown in fig

Question 10

A  3.56H  inductor is placed in series with a  12.8\Omega  resistor. An emf of  3.24V  is then suddenly applied across the  RL  combination. At  0.278s  after the emf is applied what is the rate at which energy is being delivered by the battery?

Accepted Answer

{ 	au  }_{ L }=\cfrac { L }{ R } =\cfrac { 3.56 }{ 12.8 } =0.278s { i }_{ 0 }=\cfrac { V }{ R } =\cfrac { 3.24 }{ 12.8 } =0.253A After one time constant ( t=0.278s={	au}_{C} ) \quad i=\left( 1-\cfrac { 1 }{ e }  ight) { i }_{ 0 }=0.63{ i }_{ 0 }=0.16A Power supplied by battery  =Ei P=(3.24	imes (0.16)=0.518W

Question 11

In column I, some circuits are given. In all the circuits except in (i), switch S remains closed for a long time and then it is opened at  t=0 ; while for (i), the situation is reversed. Column II tells something about the circuit quantities. Match the entries of column I with the entries of column II.

Accepted Answer

(A) if t0, I=I_0{ e }^{ -\dfrac { Rt }{ L }  }  at steady state  I=0 energy stored in the inductor= \dfrac{1}{2}LI^{2} =0(B) Energy losses through R as heat form and finally  I=0 = \dfrac{1}{2}LI^{2} =0(C) if tE (D) same as (C) but here when S-open,no resistance to dissipate energy in the inductor.

Question 12

Column I shows the cylindrical region of radius  r  where a downward magnetic field  \vec { B }   exists, where  \vec { B }   is increasing at the rate of  dB/dt . A rod  PQ  is placed in different situation as shown. Match the column I with the current statement in column II regarding the induced emf in rod.

Accepted Answer

Note:-  e=\dfrac { -d\phi  }{ dt } =\dfrac { -d }{ dt } BA\cos { 	heta  }  (A)  \Rightarrow   Induced emf { e }_{ 1 }<\dfrac { dB }{ dt } 	imes \dfrac { { r }^{ 2 }	heta  }{ 2 }  Since the rod is inside and its does'not cover the entire sector.(B) Here as Q is nearer to the field thus Q is positive w.r.t P.(C) As the magnetic field is felt only in the area of the sector having angle  '	heta' . 	herefore  Induced emf  e=\dfrac { 1 }{ 2 } { r }^{ 2 }	heta \dfrac { dB }{ dt }  (D) Similer to (B), point Q lies inside the field, thus Q is positive w.r.t P.

Question 13

A frame ABCD is rotating with an angular velocity

ω

about an axis passing through the point O perpendicular to the plane of paper as shown in fig 4.212. A uniform magnetic field

\overset{ˉ}{B}

is applies into the plane of paper in the region as shown below. Match the following.

Accepted Answer

As we know  e=\dfrac{1}{2} B\omega l^2 then,i.  e_{OA}=\dfrac {1}{2} B \omega (OA)^2=\dfrac {1}{2} B \omega (\sqrt {2}L)^2=B \omega L^2 ii.  e_{OD}=\dfrac {1}{2} B \omega (OD)^2=\dfrac {1}{2} B \omega (\sqrt {2}L)^2=B \omega L^2 iii. e_{OC}=\dfrac {1}{2} B \omega L^2  or  e_{CD}=E_{OD}-E_{OC}=\dfrac {1}{2} B \omega L^2  iv.  e_A-e_O=B\omega L^2 e_D-e_O=B\omega L^2 e_A-e_D=0

Question 14

Question 15

Charge (in  \mu C) on capacitor as a function of time is

Accepted Answer

Since charge is in phase difference with current,And thus the equation is in the form of  q = q_{max} \sin (\omega t + \phi) Now, capacitor is fully charged initially with  10\mu C , which is the rms value,Thus peak value   q_{max} = 10 \sqrt 2 Also, Initial phase difference is given by, 	an \phi = \dfrac{X_C}{X_L} = \dfrac{1}{\omega ^2 	imes LC} = 1 	herefore \phi = \dfrac{\pi}{4} Also,  \omega = \sqrt {\dfrac{1}{LC}} = 10^4

Question 16

Maximum current in the inductor coil at any time  t > 0  is :

Accepted Answer

Maximum current ...........................................After closing  S_1, S_2 \ and S_3 : At any Time  t  :  \dfrac{d^2 q}{dt^2}=-\omega^2 q \Rightarrow q = Q_{max} \sin (\omega t + \phi ) The t=0 Capacitor has charge  Q=CV = 10^{-6} 	imes 10 \ \mu C  and current in inductor will be  i=\dfrac{10}{100}=\dfrac{1}{10}\  A Using conservation of energy,  \dfrac{1}{2}(10 mH)(\dfrac{1}{10})^2 + \dfrac{(10\mu C)^2}{2.(1 \mu F)}=    \dfrac{1}{2}(10mH)I_{max}^2 I_{max}=\dfrac{1}{5\sqrt2}A At   t=0, q=10  \ \mu C   so  \phi = \dfrac{\pi}{4} Also  \dfrac{di}{dt}=q So  \dfrac{di}{dt}  will become maximum when charge becomes maximum.

Electromagnetic Induction