Saved Bookmarks
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1201. |
Consider the following statements: (a)An emf can be induced by moving a conductor in a magnetic field. (b)An emf can be induced by changing the magnetic field.A. Both A and B are trueB. A is true but B is falseC. B is true but A is falseD. Both A and B are false |
| Answer» Correct Answer - A | |
| 1202. |
In a coil of resistance `100 Omega` , a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is A. 275 WbB. 200WbC. 225WbD. 250Wb |
| Answer» Correct Answer - D | |
| 1203. |
Mutual induction is the produced of induced emf in a coil due to the change of current in theA. same coilB. neighbouring coilC. both a and bD. neigher a nor b |
| Answer» Correct Answer - B | |
| 1204. |
If the radius of a coil is changing at the rate of `10^(-2)` unit in a normal magnetic field of `10^(-3)` units, the induced emf is `1muV`. What the final radius of the coil?A. 1.6 cmB. 16 cmC. 12 cmD. 1.2 cm |
| Answer» Correct Answer - A | |
| 1205. |
The core of transformer is laminated so thatA. Ratio of voltage in the primary and secondary may be increasedB. Energy losses due to eddy currents may be minimisedC. Weights of transformer may be reducedD. Rusting may be prevented |
| Answer» Correct Answer - B | |
| 1206. |
A long solenoid having 1000 turns per cm carries an alternating current of peak value 1 A. A search coil having a cross-sectional area of `1xx10^(-4) m^(2)` and 20 turns is kept in the solenoid so that its plane is perpendicular to the axis of the solenoid. The search coil registers a peak voltage of `2.5xx10^(-2)` V. Find the frequency of the current in the solenoid. |
|
Answer» Correct Answer - `15.8 s^(-1)` |
|
| 1207. |
A circular coil with a cross-sectional area of `4cm^(2)` has 10 turns. It is placed at the center of a long solenoid that has 15 turns/cm and a cross sectional area of `10cm^(2)`, shown in the figure. The axis of the coil conicides with the axis of the solenoid. What is their mutual inductance? A. `7.54muH`B. `8.54muH`C. `9.54muH`D. `10.54muH` |
|
Answer» Correct Answer - A Let us refer to the coil as circuit 1 and the soleniod circuite 2. The field in the central region of the solenoid is uniform, so the flux through the coil is `phi_(12)=B_(A)A_(1)=(mu_(0)n_(2)I_(2))A_(1)` where `n_(2)=N_(2)//l=1500` turn/m The mutual inductance si `M=(N_(2)phi_(12))/(I_(2))=mu_(0)n_(2)N_(1)A_(1)` `=(4pixx10^(-7)Tm A^(-1))(1500m^(-1))(10)(4xx10^(-4)m^(2))` `7.54xx10^(-6)H=7.54 muH` |
|
| 1208. |
If a coil of `40` turns and area `4.0 cm^(2)` is suddenly remove from a magnetic field, it is observed that a charge of `2.0xx10^(-4)C` flows into the coil. If the resistance of the coil is `80Omega`, the magnetic flux density in `Wb//m^(2)` isA. 0.5B. `1.0`C. 1.5D. `2.0` |
| Answer» Correct Answer - B | |
| 1209. |
If a coil of `40` turns and area `4.0 cm^(2)` is suddenly remove from a magnetic field, it is observed that a charge of `2.0xx10^(-4)C` flows into the coil. If the resistance of the coil is `80Omega`, the magnetic flux density in `Wb//m^(2)` isA. `0.5`B. `1.0`C. `1.5`D. `2.0` |
|
Answer» Correct Answer - B `DeltaQ=(Deltavarphi)/(R )=(nxxBA)/(R )` `impliesB=(DeltaQ.R)/(nA)=(2xx10^(-4)xx80)/(40xx4xx10^(-4))=1 Wb//m^(2)` |
|
| 1210. |
A solenoid has a cross sectional area of `6.0xx10^(-4) m^(2)`, consists of `400` turns per meter, and carries a current of `0.40 A`.`A 10` turn coil is wrapped tightly around the circumference of the solenoid.The ends of the coil are connected to a `1.5 Omega` resistor.Suddenly, a switch is opened and the current in the solenoid dies to zero in a time `0.050 s`,Find the average current in the coil during the time. |
|
Answer» Correct Answer - A `ltigt = (ltepsilongt)/R=(BA)/(Rt)=((mu_(0)nI)AN)/(Rt)=(4pixx10^(-7)xx400xx0.40xx6xx10^(-4)xx10)/(1.5xx0.050)` |
|
| 1211. |
If a coil of `40` turns and area `4.0 cm^(2)` is suddenly remove from a magnetic field, it is observed that a charge of `2.0xx10^(-4)C` flows into the coil. If the resistance of the coil is `80Omega`, the magnetic flux density in `Wb//m^(2)` isA. 0.5B. 1C. 1.5D. 2 |
|
Answer» Correct Answer - B Charge, `DeltaQ=(Deltaphi)/(R)=(nxxBA)/(R)` `rArr` Magnetic flux density, `B=(DeltaQ.R)/(nA)=(2xx10^(-4)xx80)/(40xx4xx10^(-4))=1"Wb/m"^(2)` |
|
| 1212. |
An infinitely long cylinder is kept parallel to an uniform magnetic field B directed along positive z-axis. The direction of induced current as seen from the z-axis will beA. (a) zeroB. (b) anticlockwise of the positive z-axisC. ( c) clockwise ogf the positive z-axissD. (d) along with magnetic field |
|
Answer» Correct Answer - A For a current to induce in the cylindrical conducting rod, the cylindrical rod should cut magnetic lines of force which will happen only when the cylindrical conducting rod is moving. Since conducting rod is at rest, no current will be electric field rod and a current will get induced. But since the magnet field is constant, no current will be induced. |
|
| 1213. |
An infinitely long cylinder is kept parallel to an uniform magnetic field B directed along positive z-axis. The direction of induced current as seen from the z-axis will beA. clockwise of the positive Z-axisB. anti positive Z-axis clockwise of the positive z-axisC. zeroD. along the magnetic field |
|
Answer» Correct Answer - C Since, the magnetic field is uniform, therefore there will be no change in flux, hence no current will be induced. |
|
| 1214. |
A coil of inductance L is carrying a steady current l. what is the nature of its stored energy ?A. MagneticB. ElectricalC. Both magnetic and electricalD. Heat |
|
Answer» Correct Answer - A Energy is stored in an inductor in the from of magnetic potential energy. |
|
| 1215. |
When a coil is rotated in a magnetic field, with steady speed, thenA. no e.m.f. is inducedB. a periodic e.m.f. is inducedC. unidirectional e.m.f. is inducedD. multidirectional e.m.f. is induced |
| Answer» Correct Answer - B | |
| 1216. |
In the circuit shown , what is the energy stored in the coil at steady state? A. 21.3 JB. 42.6 JC. zeroD. 213 J |
|
Answer» Correct Answer - C Four resistance from a balanced Wheatstone bridge. Therefore, energy stored in the coil is zero. |
|
| 1217. |
In an LCR series circuit, the voltage across each of the components L,C and R is 20 V. The voltage across the L-C combination will beA. 20 VB. 40 VC. zeroD. `20sqrt(2)V` |
|
Answer» Correct Answer - C The voltage across L and C combination wil be zero. |
|
| 1218. |
In an LCR series a.c. circuit, the voltage across each of the components L,C, and R is 60 V. What is the voltage across the LC combination ?A. 60 VB. 120 VC. zero VD. `60/(sqrt2)V` |
|
Answer» Correct Answer - C In LCR series circuit, the voltage across L leads the current by `90^(@)` and the voltage across C lags behind the current by `90^(@)`. Hence the voltage across the LC combination is zero. |
|
| 1219. |
A current `I=100sqrt(2) cos (omega t - phi)` is passed through a D.C. ammeter. The ammeter will readA. a. `100sqrt(2)`B. b. `100 A`C. c. `100/(sqrt2) A`D. d. zero |
|
Answer» Correct Answer - D D.C. ammeter will show zero deflection. |
|
| 1220. |
In an LCR series a.c. Circuit the voltage across each of hte components L,C and R is 50V. The voltage across the LC combination will beA. 100VB. `50sqrt(2)V`C. 50VD. 0V(zero) |
|
Answer» Correct Answer - D (d) Since the phase difference between L & C is `(pi)`, `:. Net voltage different across LC=50-50=0. |
|
| 1221. |
Alternating current can not be measured by D.C. Ammeter becauseA. Average value of current for current for complete cycle is zeroB. AC changes directionC. A.C can not pass through D.C Ammeter will get damaged.D. |
|
Answer» Correct Answer - A (a) D.C. Ammeter measure average current in AC current, average current is zero for complete cycle. Hence reading will be zero. |
|
| 1222. |
In the circuit shown Fig. when frequency of supply is doubled, how should the values of L and C be changed so that glow in the bulb remains unchanged ? |
|
Answer» Current in RLC circuit is `I_(v) = (E_(v))/(Z) = (E_(v))/(sqrt(R^(2) + (2 pi v L - (1)/(2 pi v C))^(2)` The glow of bulb will remain unchanged when `I_(v)` remains constant. When v is doubled, L should be halved and C shoud also be halved. |
|
| 1223. |
Can a capacitor of suitable capacitance be use dto control a.c in place of the choke coil ? |
| Answer» Yes, a capacitor can be used, as it also involves no dissipation of energy. The capactative reactance is `X_(C ) = 1//omega C`. | |
| 1224. |
What is the resistance to be connected in series with a condenser of a capacity `5muF` so that the phase difference between the current and the applied voltage is `45^(@)` when the angular frequency of applied voltage is 400 rad/s ?A. `250Omega`B. `400Omega`C. `500Omega`D. `600Omega` |
|
Answer» Correct Answer - C `tanphi=(1)/(omegaCR)` `R=(1)/(omegaC tanphi)=(1)/(400xx5xx10^(-6)xxtan45)` `=500Omega` |
|
| 1225. |
Modern day generators produce electric power as high as 500 MW. How many 100 watt bulb can one lit with this generator? |
| Answer» Correct Answer - 5 million. | |
| 1226. |
In an LCR circuit, inductive reactance and capacitive reactance was found to be equal. The resistance was found to be `20Omega`. The probable impedance of the combination isA. zeroB. `20Omega`C. `40sqrt(2)Omega`D. `400Omega` |
|
Answer» Correct Answer - B If `X_(L)=X_(C)` then Z=R `therefore Z=20Omega` |
|
| 1227. |
The rms value of an ac of 50Hz is 10A. The time taken by an alternating current in reaching from zero to maximum value and the peak value will beA. 0.02 s and 14.14 AB. 0.01 s and 7.07 AC. 0.005 s and 7.07 AD. 0.005 s and 14.14 A |
|
Answer» Correct Answer - D `theta=omegat=2pift` `therefore t=(theta)/(2pif)` `=(pi)/(2xx2pixx50)=0.005` sec. `I_(0)=sqrt(2)I_(rms)=1.414xx10=14.14A` |
|
| 1228. |
A triangular wire frame (each side =2m) is placed in a region of time variant magnetic field `dB//dt = (sqrt3) T//s`. The magnetic field is perpendicular to the plane of the triangle and its centre coincides with the centre of triangle. The base of the triangle AB has a resistance `1(Omega)` while the other two sides have resistance `2(Omega)` each. The magnitude of potential difference between the points A and B will be A. 0.4VB. 0.6VC. 1.2VD. None |
|
Answer» Correct Answer - B |
|
| 1229. |
A uniform magnetic field is restricted within a region of radius r. The magnetic field changes with time at a rate `(dB)/(dt)`. Loop 1 of radius R `gt` r encloses the region r and loop 2 of radius R is outside the region of magnetic field as shown in figure. Then, the emf generated is A. Zero in loop 1 and zero in loop 2B. `-(dB)/(dt)pir^(2)` in loop 1 and `-(dB)/(dt)pi^(2)` in loop 2C. `-(dB)/(dt)piR^(2)` in loop 1 and zero in loop 2D. `-(dB)/(dt)pir^(2)` in loop 1 and zero in loop 2 |
|
Answer» Correct Answer - C Induced emf in the region is given by `|e|=(dhpi)/(dt)` where, `phi=BA=pir^(2)B` `rArr" "(dphi_(1))/(dt)-pir^(2)(dB)/(dt)` Rate of change if magnetic flux associated with loop l, `e_(1)=-(dphi_(1))/(dt)=-pir^(2)(dB)/(dt)` Similarly, `e_(2)` = emf associated with loop 2 `=-(dphi_(2))/(dt)=0" "(because phi_(2)=0)` |
|
| 1230. |
Are eddy currents useful or harmful ? |
| Answer» They are both, useful and harmful. | |
| 1231. |
cylindrical volume of radius R has a uniform axial magnetic field B, which is increasing at a rate of `(dB)/(Dt) = alpha Ts^(-1)`. A chord (AB) of the circular cross section of the cylindrical region has length L.Calculate the line integral of induced electric field `(int_(A)^(B) vec(E) vec(dl))` as one moves along the chord from A to B. Try to find the answer without actually performing the integration. Is the value of integral same if one moves along the arc from A to B ? |
|
Answer» Correct Answer - `(L alpha)/(4) sqrt(4R^(2)-L^(2))` |
|
| 1232. |
A current carrying ring is placed in a horizontal plane. A charged particle is dropped along the axis of the ring to fall under the influence of gravityA. The current in the ring may increaseB. The current in the ring may decreaseC. The velocity of the particle will increase till it reaches the centre of the ring.D. The acceleration of the particle will decrease continuosly till it reaches the centre of the ring. |
|
Answer» Correct Answer - A |
|
| 1233. |
Eddy currents are produced in a matterial when it isA. A metal body is kep in a time varying magnetic fieldB. A metal body is kept in the steady magnetic fieldC. A circular coil is placed in a magnetic fieldD. Through a circular coil current is passed. |
|
Answer» Correct Answer - B |
|
| 1234. |
There is a horizontal cylindrical uniform but time-varying nagnetic field increasing at a constant rate `dB//dt` as shown in Fig. 3.173. A charged particle having charge `q` and mass `m` is kept in euilibrium, at the top of a spring of spring constant `K`, in such a way that it is on the horizontal line passing through the center of the magnetic field as shown in the figure. The compression in the spring will be A. (a) `(1)/(K)[mg - (qR^(2))/(2l)(dB)/(dt)]`B. (b) `(1)/(K)[mg + (qR^(2))/(l)(dB)/(dt)]`C. ( c) `(1)/(K)[mg + (2qR^(2))/(l)(dB)/(dt)]`D. (d) `(1)/(K)[mg + (qR^(2))/(2l)(dB)/(dt)]` |
|
Answer» Correct Answer - D (d) `E2pil = pir^(2)((dB)/(dt)), E = (R^(2))/(2l)((dB)/(dt))` `qE + mg = Kx` `rarr` `x = (qR^(2))/(K2l)((dB)/(dt)) + (mg)/(K), x = (1)/(K)[(mg + (qR^(2))/(2l)(dB)/(dt)]` |
|
| 1235. |
A magnetic field induction is changing in magnitude in a region at a constant rate `dB//dt`. A given mass `m` of copper drawn into a wire and formed into a loop is placed perpendicular to the field. If the values of specific resistance and density of copper are `rho` and `sigma` respectively, then the current in the loop is given by :A. (a) `(m)/(2pirhod)(dB)/(dt)`B. (b) `(m)/(4pia^(2)r)(dB)/(dt)`C. (c) `(m)/(4piad)(dB)/(dt)`D. (d) `(m)/(4pirhod)(dB)/(dt)` |
|
Answer» Correct Answer - D (d) `xi = (dphi)/(dt) `rarr` I = (xi)/(R )(d)/(dA)(BA) = (A)/(R )(dB)/(dt)` where `pir^(2)` = area of the loop of radius `r` and `R` = resistance of the loop of length `(2pir)` and area of cross section `pia^(2)`. `R = (rhol)/(pia^(2)) = (rho(2pir))/(pia^(2))` Further mass of wire is `m = (pia^(2))(2pir)(d)` `i = ((pia^(2))(pir^(2)))/(rho(2pir))(dB)/(dt)` `i = ((pia^(2))(pir^(2)))/(4pirho)(dB)/(dt) rarr i = (m)/(4pirhod)(dB)/(dt)` |
|
| 1236. |
A loop of wire enclosing an area `S` is placed in a region where the magnetic field is perpendicular to the plane. The magnetic field `B` varies with time according to the expression `B=B_0e^(-at)` where a is some constant. That is, at `t= 0`. The field is `B_0` and for `t gt 0`, the field decreases exponentially. Find the induced emf in the loop as a function of time. |
|
Answer» Correct Answer - A::B `phi=BS=B_2Se^(-at)` induced emf`=|(dphi)/(dt)|=aB_0Se^(-at)` |
|
| 1237. |
The figure shows as circular region of radius `R` occupied by a time varying mgnetic field `B(t)` such that `(dB)/(dt)lt0`. The magnitude of induced electric field at the point `P` at a distance `r lt R` is A. decreasing with `r`B. increasing with `r`C. not varying with `r`D. varying as `r^2` |
|
Answer» Correct Answer - B `El=(dphi)/(dt)=S (dB)/(dt)` `:. E(2pir)=pir^2(dB)/(dt)` or `E=r/2 (dB)/(dt)` or `Epropr` |
|
| 1238. |
A solenoid with an iron core and a bulb are connected to a d.c. source. How does the brightness of the bulb change when iron core is removed from the solenoid ? |
| Answer» The brightness of bulb remains unchanged because the solenoid does not offer any reactance `(X_(L) = 2 pi v L)` to d.c. source `(v = 0)`. | |
| 1239. |
In a very long solenoid of radius `R`, if the magnetic field chabges at the rate of `dB//dt.AB = BC` The induced emf for the trianglar circuit `ABC` shown in Fig. 3.196 is A. (a) `R^(2)((dB)/(dt))`B. (b) `4R^(2)((dB)/(dt))`C. ( c) `(1)/(2)R^(2)((dB)/(dt))`D. (d) `2R^(2)((dB)/(dt))` |
|
Answer» Correct Answer - C ( c)`|E| = (dphi)/(dt) = A(dB)/(dt) = 2[(1)/(2)R R] (dB)/(dt) = R^(2)(dB)/(dt)` In loop `ABC`, emf induced due to branch `AC` is zero and contribution of emf due to `AB and BC` are equal, hence contribution of emf for the branch is `|E|_(AB) = (R^(2))/(2)((dB)/(dt))` |
|
| 1240. |
In a very long solenoid of radius `R`, if the magnetic field chabges at the rate of `dB//dt.AB = BC` The induced emf for the trianglar circuit `ABC` shown in Fig. 3.196 is A. (a) `R^(2)((dB)/(dt))`B. (b) `4R^(2)((dB)/(dt))`C. ( c) `(1)/(2)R^(2)((dB)/(dt))`D. (d) `2R^(2)((dB)/(dt))` |
|
Answer» Correct Answer - A (a)`|E| = (dphi)/(dt) = A(dB)/(dt) = 2[(1)/(2)R R] (dB)/(dt) = R^(2)(dB)/(dt)` In loop `ABC`, emf induced due to branch `AC` is zero and contribution of emf due to `AB and BC` are equal, hence contribution of emf for the branch is `|E|_(AB) = (R^(2))/(2)((dB)/(dt))` |
|
| 1241. |
In figure, a lamp P is in series with an iron-core inductor L. When the switch S is closed, the brightness of the lamp rises relatively slowly to its full brightness than it would do without the inductor. This is due to A. the low resistance of PB. the induced EMF in LC. the low resistance of LD. the high voltage of the battery B |
|
Answer» Correct Answer - B |
|
| 1242. |
(a) Find the emf induced in the coil if it were positioned such that its plane contains the axis of the solenoid (b) In the above problem (example 3) what is the emf induced when current is reduced to zero and then increases again to 1.5 A in reverse direction in 25 ms. |
|
Answer» (a) No emf is induced in the loop even though the magnetic field in space is changing. This is because no flux is linked with the coil. (b) 150 mV |
|
| 1243. |
A coil formed by wrapping 50 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of `30^@`. with the direction of the field. When the magnetic field is increased uniformly from `200 muT` to `600muT` in `0.4 s`, an emf of magnitude 80.0 mV is induced in the coil. What is the total length of the wire? |
|
Answer» Correct Answer - B `e=(N/_phi)/(/_ )=((NS/_B)/(/_ ))cos30^@` `:. S=((e/_ )/(N/_B))sec30^@` `=[((80xx10^-3)(0.4))/((50)(400x10^-6))][2/sqrt3]=1.85m` `:.` Side of square `=1.36m` Total length of wire `=50(4xx1.36)` `=272m` |
|
| 1244. |
A coil formed by wrapping 50 truns of wire in the shape of square is positioned in a magnetic field so that the normal to plane of the coil makes an angle of `60^(@)` , with the direction of the field. When the magnetic field is increased uniformly from `200muT` to `600muT` in `0.4sec` , an emf of magnitude `I00mV` is induced in the coil. What is the total length of the wire ? |
|
Answer» `N=50` , `theta=60^(@)` , `bar(e)=100xx10^(-3)V` `(DeltaB)/(Deltat)=((600-200)xx10^(-6))/(0.4)=10^(-3)T//sec` Let side of square be a `phi=NBAcostheta=Nba^(2)cos60^(@)=(1)/(2)Nba^(2)` `bar(e)=(Deltaphi)/(Deltat)=(1)/(2)Na^(2)(DeltaB)/(Deltat)=(1)/(2)xx50xxa^(2)xx10^(-3)` `100xx10^(-3)=25xx10^(-3)a^(2)` implies `a=2m` Total length of wire `=N(4a)=50xx4xx2=400m` |
|
| 1245. |
A lamp in a circuit consisting of a coil of large number of truns and a battery does not ligth upto full brilliance instantly on switching on the circuit ? |
| Answer» When the circuit is switched on. Current increase through the lamp and also through the coil. An induced e.m.f. developes in the coil which oppose the growth of current. As the current takes time to maximum value, the lamp does not light instantly upto full brilliance. | |
| 1246. |
Asseration: When two coils are wound on each other, the mutual induction between the coils is maximum. Reason: Mutual induction does not depend on the orientation of the coils.A. If both asseration and reason are true and reason is the correct explanation of assertion.B. If both asseration and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
|
Answer» Correct Answer - C Coefficient of coupling between them `M=K^(2).L_(1)L_(2)` when `2` coils are wound on each other, the coefficient of coupling is maximum and hence mutual inductance between the coils is maximum. |
|
| 1247. |
Asseration:Induced coil are made of copper. Reason:Induced current is more in wire having less resistance. |
|
Answer» Correct Answer - a The inductance coils made of copper will have very small ohmic resitance. Due to change in magnetic flux a large induced current will be produced in such an inductance, which will offer appericable opposition to the flow of current. |
|
| 1248. |
Asseration: The induced emf in a conducting loop of wire will be non zero when it rotates in a uniform magnetic field. Reason: The emf is induced due to change in magnetic flux.A. If both asseration and reason are true and reason is the correct explanation of assertion.B. If both asseration and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
|
Answer» Correct Answer - A As the coil rotates, the magnetic flux linked with the coil (being `vecB.vecA`) will change and emf will be induced in the loop. |
|
| 1249. |
Asseration: The magnetic flux through a loop of conducting wire of a fixed resistance changes by `Deltaphi_(B)` in a time `Deltat`. Then `Deltaphi_(B)` is proportional to the current through the loop. Reason: `I=-(Deltaphi_(B))/(R ))`A. If both asseration and reason are true and reason is the correct explanation of assertion.B. If both asseration and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
|
Answer» Correct Answer - A `I=-(Deltaphi_(B))/(R )` As `R` is constant, `IpropDeltaphi_(B). |
|
| 1250. |
The magnetic flux through a circuit of resistance `R` changes by an amount `Omegaphi` in a time `Deltat`. Then the total quantity of electric charge `Q` that passes any point in the circuit during the time `Deltat` is represent byA. `Q=(1)/(r ).(Deltaphi)/(Deltat)`B. `Q=(Deltaphi)/(R )`C. `Q=(Deltaphi)/(Deltat)`D. `Q=(R ).(Deltaphi)/(Deltat)` |
|
Answer» Correct Answer - D The emf induced in the circuit `e=(Deltavarphi)/(Deltat)` If `R` is the resistance of the circuit, then `i=(e)/(R )=(Deltavarphi)/(Rdeltat)` thus, charge passes through the circuit, `Q=ixxDeltat` `impliesQ=(Deltavarphi)/(Rdeltat)xxDeltat` `impliesQ=(Deltavarphi)/(R )` |
|