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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.
| 501. |
In the Figure shown, when the switch S is closed, one of the two bulbs glows momentarily. Which one? |
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Answer» Correct Answer - Bulb A |
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| 502. |
A solenoid of inductance `L` with resistance `r` is connected in parallel to a resistance `R`. A battery of emf `E` and of negligible internal resistance is connected across the parallel combination as shown in the figure. At time `t = 0`, switch `S` is opened, calculate (a) current through the solenoid after the switch is opened.(b) amount of heat generated in the solenoid. |
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Answer» Correct Answer - B Steady state current developed in the inductor `=E/r =i_0(say)` a. Now this current decreases to zero exponentialy through `r` and `R`. `:. i=i_0r^(-t//tau_L)` where `tau_L=L/(R+r)` Energy stored in inductor, `U_0=1/2Li_0^2=(1/2L)(E/r)^2` Now, this energy dissipates in `r` and `R` in direct ratio of resistance. `:. H_r=(r/(R+r))U_0=(E^2l)/(2r(R+r))` |
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| 503. |
A long straight wire of cross sectional radius a carries a current I. The return current is carried by an identical wire which is parallel to the first wire. The centre to centre distance between the two wires is d. Find the inductance (L) of a length x of this arrangement. Neglect magnetic flux inside the wires. |
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Answer» Correct Answer - `L = (mu_(0)x)/(pi) ln ((d-a)/(a))` |
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| 504. |
A conductuing loop having a capacitor is moving outward from the magnetic field. Which plate of the capacitor will be positive? A. plate-`A`B. plate-`B`C. plate-`A` and plate -`B` bothD. none |
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Answer» Correct Answer - A Cross (xx) linked with the loop are decreasing, so induced current in it is clockwise i.e., from `Brarr A`. Hence electrons flow plate `A` to `B` so plate `A` becomes positively charged. |
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| 505. |
For a series L C R circuit, the power loss at resonance isA. `I^(2)omega L`B. `I^(2)omega C`C. `I^(2) R`D. `(E^2)/(sqrt(R^(2)+(omegaL-1/(omega C))^(2)))` |
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Answer» Correct Answer - C At resonance, `Z = R` `:.` Powerloss is only due to R and is equal to `I^(2)R`. |
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| 506. |
A circular cpper disc of 10 cm in diameter rotates at 1800 revolution per minute about an axis through its centre and at right angles to disc. A uniform field of induction B of `1Wb^(-2)m` is perpendicular to disc. What potential difference is developed between the axis of the disc and the rim?A. `0.023V`B. `0.23V`C. `23V`D. `230V` |
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Answer» Correct Answer - B Here, `l=r=5cm=5xx10^(-2)m` `omega=2pi((1800)/(60))"rad" s^(-1)=60 pi "rad" s^(-1)`, `B=1WB m^(-2)` `epsi=(1)/(2)Bl^(2) omega=(1)/(2)xx1xx(5xx10^(-2))^(2)xx60pi=0.23V` |
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| 507. |
Find the current in the sliding rod AB (resistance = R) for the arranged shown in Fig. B is constant and is out of the paper. Paralllel wires have no resistance. `upsilon` is constant. Switch is closed at time t = 0. |
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Answer» Proceeding as in Q.29 above, `- L (dI)/(dt) + B upsilon d = IR` or `L (dI)/(dt) + IR = B upsilon d` `:. I = (B upsilon d)/(R ) + A e^(-Rt//L)` where A is an arbitrary constant. At t = 0, I = 0 `:.` From (i), `A = - (B upsilon d)/(R )` Putting this value of A in (i), we get `I = (B upsilon d)/(R ) = (B upsilon d)/(R ) e^(-Rt//L) = (B upsilon d)/(R ) (1 - e^(-Rt//L))` This is the required current in sliding rod AB. |
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| 508. |
A metallic ring of field. If z is the radius l (ring being horizontal is falling under gravity in a region haivng a magnetic field. If z is the vertical direction, the z-component of magnetic field is `B_(z) = B_(0) (1 + lambda z)`. If R the resistance of the ring and if the ring falls with a velocity `upsilon`, find the energy lost in the resistance If the ring has reached a constant velocity, use the conservation of energy to determine `upsilon` in terms of m, B, `lambda` and acceleration due to gravity g. |
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Answer» Here, `B_(z) = B_(0) (1 + lambda z)` As the metallic ring falls under gravity in this magnetic field, magnetic flux linked with the ring changes Therefore, an emf is induced in the ring. If I is induced current in the ring at any instant, then induced emf ` = IR = (d phi)/(dt) = A xx (dB_(z))/(dt) = (pi l^(2)) B_(0) lambda (dz)/(dt)` `:. I = (pi l^(2) B_(0) lambda upsilon)/(R )`, where `(dz)/(dt) = upsilon` Energy `lost//sec` in the from of heat produced in the ring `= I_(2) R = ((pi l^(2) lambda)^(2) B_(0)^(2) upsilon^(2))/(R )` This comes from rate of change of potential energy of ring `= mg (dz)/(dt) = mg upsilon` `:. ((pi l^(2) lambda)^(2) B_(0)^(2) upsilon^(2))/(R ) = mg upsilon` or `upsilon = (mg R)/((pi l^(2) lambda B_(0) )^(2))` |
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| 509. |
Two identical thin circular, metal plates are at a small separation d. They are connected by a thin conducting rod (AB) of length d. Each plate has area A. An ideal spring of stiffness k is connected to a rigid support and midpoint of rod AB as shown in the figure. Spring is made of insulating material. The system is on a smooth horizontal surface. The entire region has a uniform vertical upward magnetic field B. The discs are pulled away from the support and released. Find time period of oscillations. Mass of the two disc plus rod system is M. Neglect any eddy current. |
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Answer» Correct Answer - `T = 2pi sqrt((M+in_(0)AB^(2).d)/(k))` |
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| 510. |
A conducting rod of mass M and length L can oscillate like a pendulum in a vertical plane about point O. The lower end of the rod glides smoothly on a circular conducting arc of radius L. The circular arc is connected to point O of the rod through a capacitor of capacitance C. The entire device is kept in a uniform horizontal magnetic field B directed into the plane of the Figure. Disregard resistance of any component. The rod is deflected through a small angle `theta_(0)` from vertical position and released at time t = 0. (a) Write the deflection angle `(theta)` of the rod as a function of time t. (b) If the capacitor is replaced with a resistor what kind of motion do you expect? Give qualitative description only. |
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Answer» Correct Answer - (a) `theta = theta_(0) cos omega t` where `omega = [(6Mg)/(4ML + 3B^(2)L^(3)C)]^(1//2)` (b) Oscillations die after some time. |
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| 511. |
The number of turns in the primary coil of a transformer is `200` and the number of turns in the secondary coil is `10` if `240` volt `AC` is applied to the primary, the output from the secondary will beA. 48 VB. 24 VC. 12 VD. 6 V |
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Answer» Correct Answer - C `e_(S)=(n_(S))/(n_(P))xxe_(P)=(10)/(200)xx240=12V` |
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| 512. |
The number of turns in the primary coil of a transformer is `200` and the number of turns in the secondary coil is `1.0` if `240` volt `AC` is applied to the primary, the output from the secondary will beA. `48V`B. `24V`C. `12V`D. `6V` |
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Answer» Correct Answer - C `(V_(s))/(V_(p))=(N_(s))/(N_(p))impliesV_(s)=(N_(s))/(N_(p))xxV_(p)=(10)/(200)xx240=12` volts |
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| 513. |
In a step-down transformer input voltage is 200 volts and output voltage 5 volt. The ratio of nubmer of turns in it will be -A. `1:40`B. `40:1`C. `20:1`D. `1:20` |
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Answer» Correct Answer - B `(n_(1))/(n_(2))=200/5 =40` |
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| 514. |
A step-down transformer is connected to `2400` volts line and `80` amperes of current is found to flow in output load. The ratio of the turns in primary and secondary coil is `20:1.` if transformer efficiency is `100%`, then the current flowing in primary coil will beA. `15A`B. `50A`C. `25A`D. 12.5A` |
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Answer» Correct Answer - B `(N_(s))/(N_(p))=(i_(p))/(i_(s))or (25)/(1)=(i_(p))/(2)impliesi_(p)=50A` |
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| 515. |
In the circuit shown below, the key K is closed at t = 0. The current through the battery is A. `(VR_(1)R_(2))/(sqrt(R_(1)^(2)+R_(2)^(2))) at t=0 and (V)/(R^(2)) at t=oo`B. `(V)/(R^(2)) at t=0 and (V(R_(1)+R_(2)))/(R_(1)R_(2)) at t=oo`C. `(V)/(R^(2)) at t=0 and (VR_(1)R_(2))/(sqrt(R_(1)^(2)+R_(2)^(2))) at t=oo`D. `(v(R_(1)+R_(2)))/(R_(1)R_(2)) at t=0 and (V)/(R_2) at t=oo` |
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Answer» Correct Answer - B (c) At t-0, no current will flow through L and `(R_(1))` `:. Current through battery `(V)/(R^2)` At `t=(oo)`, effective resistnace, (R_eff)=(R_(1)R_(2))/(R_(1)+R_(2))` `:. Current through battery `(V)/R_(eff))=(V(R_(1)=R_(2))/(R_(1)R_(2))`. |
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| 516. |
When a magnet is moved with its south pole towards the coil, then the nearer face of the coil behaves as aA. N-poleB. S-poleC. `+ve` chargeD. `-ve` charge |
| Answer» Correct Answer - B | |
| 517. |
When a magnetic is moved with its N pole away from the coil, then the nearer face of the coil behaves as aA. N-poleB. S-poleC. `+ve` chargeD. `-ve` charge |
| Answer» Correct Answer - B | |
| 518. |
Induced emf can be produced byA. moving a magnet near a circuitB. moving a circuit near a magnetC. changing the current in one circuit near the otherD. all of these |
| Answer» Correct Answer - D | |
| 519. |
The direction of the induced `e.m.f.` is determined byA. right hand ruleB. right hand screw ruleC. Flemings right hand ruleD. Flemings left hand rule |
| Answer» Correct Answer - C | |
| 520. |
Which of the following phenomenon makes use of electromagnetic induction?A. Magnetising an iron piece with a bar magnetB. Magnetising a soft iron by placing it inside a current carrying solenoidC. charging a storage batteryD. Generation of hydro electricity |
| Answer» Correct Answer - D | |
| 521. |
STATEMENT - 1 : In a series L-C-R circuit, it is found that voltage leads current. When the capacitance of the circuit is increased, the power consumption will increase. and STATEMENT - 2 : The average power consumption at resonance is independent of inductor of capacitor.A. Statement-1 is True, Statement-2, is True, Statement-2 is a correct explanation for Statement-8B. Statement-1 is True, Statement-2, is True, Statement-2 is NOT a correct explanation for Statement-8C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - D | |
| 522. |
Can and induced electric field exist at a point in space, when no magentic field is paresent at that point. |
| Answer» Yes,Electric fields can exist without a magnetic field - consider a stationary point charge. Magnetic fields cannot exist without any E field component because there are no magnetic monopoles. ... A changing electric field creates a magnetic field | |
| 523. |
Can an inductor have potential difference across it ends even if does not store any energy. |
| Answer» Yes, when `i=0` but `(dl)/(dt) ne 0` | |
| 524. |
A uniform magnetic field, `B=B_(0)t` (where `B_(0)` is a positive constant) , fills a cylinderical volume of radius R, then the potential difference in the conducting rod PQ due to electorstatic field is : A. `B_(0)lsqrt(R^(2)+l^(2))`B. `B_(0)lsqrt(R^(2)-(l^(2))/(4))`C. `B_(0)lsqrt(R^(2)-l^(2))`D. `B_(0)Rsqrt(R^(2)-l^(2))` |
| Answer» Correct Answer - C | |
| 525. |
A conducting disc of radius `R` is rolling without sliding on a horizontal surface with a constant velocity `v`.A uniform magnetic field of strength `B` is applied normal to the plane of the disc.Find the `EMF` induced between (at this moment ) (a)`P&Q` , (b)`P&C` , (c )`Q&C` (`C` is centre `P&Q` are opposite points on vertical diameter of the disc) |
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Answer» Correct Answer - A::B::C (a)`epsilon_(PQ)=1/2Bomega(2R)^(2) =1/2B(v/R)(2R)^(2)=2BvR` (b)`epsilon_(PC)=1/2BomegaR^(2) =1/2B(v/R)R^(2)=(BvR)/2` ( c)`epsilon_(QC)=2Bv/R-(BvR)/2=3/2vBR` |
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| 526. |
A conducting disc of radius `R` is placed in a uniform and constant magnetic field `B` parallel to the axis of the disc.With what angular speed should the disc be rotated about its axis such that no electric field develops in the disc (the electric charge and mass are `e` and `m`) A. `(eB)/(2m)`B. `(eB)/(m)`C. `(2pim)/(eB)`D. `(pim)/(eB)` |
| Answer» Correct Answer - B | |
| 527. |
The self inductance of a coil is a measure ofA. electrical inertiaB. electrical frictionC. induced emf and the magnetic fluxD. induced current |
| Answer» Correct Answer - A | |
| 528. |
When the current in a certain inductor coil is `5.0 A` and is increasing at the rate of `10.0 A//s`,the magnitude of potential difference across the coil is `140 V`.When the current is `5.0 A` and decreasing at the rate of `10.0 A//s`,the potential difference is `60 V`.The self inductance of the coil is :A. 2HB. 4HC. 10HD. 12H |
| Answer» Correct Answer - B | |
| 529. |
One henry is equal toA. 1 weber/ampereB. 1 weber/voltC. 1 weber ampereD. 1 weber volt |
| Answer» Correct Answer - A | |
| 530. |
In SI, Henry is the unit ofA. resistanceB. capacityC. inductanceD. current |
| Answer» Correct Answer - C | |
| 531. |
Assertion An induced emf of 2 V is developed in a circular loop, if current in the loop is changed at a rate of 4 `As^(-1)` .IF 4 A of current is passed through this loop, then flux linked with this coil will be 2 Wb. Reason Flux linked with the coil is `|phi|=|(e)/(di//dt)|i`A. If both Assertion and Reason are corrent and Reason is the corrent explanation of Assertion.B. If both Assertion and Reason are correct 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 ture. |
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Answer» Correct Answer - A `L=(-e)/(di//dt)rArrL=2//4=1//2` On putting values, we get `phi=LirArr phi=(1)/(2)xx4=2Wb` |
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| 532. |
Assertion If current passing through a circular loop is doubled, then magnetic flux linked with the circular loop will also become two times. Reason No flux will link through the coil by its own current.A. If both Assertion and Reason are corrent and Reason is the corrent explanation of Assertion.B. If both Assertion and Reason are correct 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 ture. |
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Answer» Correct Answer - C `phi=Li" "rArr" "phi prop i` |
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| 533. |
Assertion If a straight wire is moved in a magnetic field, no emf will be induced across its two ends as the circuit is not closed. Reason Since, the circuit is not closed, induced current will be zero.A. If both Assertion and Reason are corrent and Reason is the corrent explanation of Assertion.B. If both Assertion and Reason are correct 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 ture. |
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Answer» Correct Answer - D IF a conducting rod of length l moving with a uniform velocity v perpendicular to a uniform magnetic field B, then induced emf will be e = Bvl. |
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| 534. |
Assertion If a loop is placed in a non-uniform (with respect to position) magnetic field, then induced emf is produced in the loop. Reason In a non-uniform magnetic field, magnetic flux passing through the loop will change. Therefore, induced emf is produced.A. If both Assertion and Reason are corrent and Reason is the corrent explanation of Assertion.B. If both Assertion and Reason are correct 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 ture. |
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Answer» Correct Answer - D In non-uniform magnetic field, magnetic field, magnetic flux will be obtained by integration, but it will not vary with time. |
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| 535. |
Find the `emf`induced in the rod in the following cases.The figures are self explanatory. |
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Answer» (a) here `vecv||vecB " so " vecv xxx vecB=0 " emf"=vecl*(vecvxxvecB)=0` (b) here `vecv||vecl " so emf"=vecl*(vecvxxvecB)=0` here `vecB||vecl " so emf"=vecl*(vecvxxvecB)=0` |
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| 536. |
A metal rod of length L is placed normal to a magnetic field and rotated through one end of rod in circular path with frequency f. The potential difference between it ends will be-A. `piL^(2)Bf`B. `BL//f`C. `piL^(2)B//f`D. fBL |
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Answer» Correct Answer - A `e=1/2 B omegaL^(2) =1/2 B 2 piLf L^(2) =piL^(2) Bf` |
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| 537. |
The identical loops of copper and aluminium are moving with the same speed in a magnetic field. Which of the following statements true?A. Induced emf and induced current are same in both loopsB. Induced emf remains same, induced current changes in both loopsC. Induced emf changes but induced current remains same in both loopsD. Induced emf will be more in aluminium loop |
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Answer» Correct Answer - B Induced emf in both the loops will be same as `e=(-dphi)/(dt)=B omega sin omega t` Sicne, B and `omega` are same for both the cases, induced emf will be same. However, induced current is given by `I=(e)/(R)` So, it will depend upon the value of resistance, lower the resistance higher will be the current. Therefore, induced current is different in both the loops. |
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| 538. |
A straight conductor 0.1 m long moves in a uniform magnetic field 0.1 T. The velocity of the conductor is `15ms^(-1)` and is directed perpendicular to the field. The emf induced between the two ends of the conductor isA. 0.10 VB. 0.15 VC. 1.50 VD. 15 V |
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Answer» Correct Answer - B Given, length of conductor, l = 0.1m Magnetic field, B = 0.1 T Veocity of conductor, `v=15ms^(-1)` The angle between v and B is `90^(@)`. When v and B are mutually perpendicular, then emf (induced) is given by `epsilon=vBl=15xx0.1xx0.1=(15)/(100)=0.15V` |
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| 539. |
A straight conductor 1 meter long moves a right angles to both, its length and a uniform magnetic field. If the speed of the conductor is `2.0 ms^(-1)` and the strength of the magnetic field is `10^(4)` gauss, find the value of induced emf in volt. |
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Answer» Correct Answer - 2 V |
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| 540. |
If in a triode valve amplification factor is 20 and plate resistance is `10 kOmega` , then its mutual conductance isA. 2 milli mhoB. 20 milli mhoC. (1/2) milli mhoD. 200 milli mho |
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Answer» Correct Answer - A Amplificantion factor `mu= 20` Plante resistance `R_(p)= 10 K Omega = 10 xx 10 ^(-3) Omega` `therefore` Mutual conductance `g_(m)= (mu)/(R_(p))` ` = (20)/( 10 xx 10^(3))` `= 2 xx 10 ^(3) " mho or 2 milli mho "` |
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| 541. |
A coil of area `100 cm^(2)` has `500` turns. Magnetic field of `0.1 "weber"//"metre"^(2)` is perpendicular to the coil. The field is reduced to zero in `0.1` second. The induced `e.m.f.` in the coil isA. `1 V`B. `5 V`C. `50 V`D. zero |
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Answer» Correct Answer - B `e=-(N(B_(2)-B_(2))A costheta)/(Deltat)` `=-(500xx(0-0.1)xx100xx10^(-4)cos0)/(0.1)=5V` |
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| 542. |
A coil of area `100 cm^(2)` has `500` turns. Magnetic field of `0.1 "weber"//"metre"^(2)` is perpendicular to the coil. The field is reduced to zero in `0.1` second. The induced `e.m.f.` in the coil isA. `1.77` voltsB. `17.7` voltC. `177` voltsD. `0.177` volts |
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Answer» Correct Answer - B `e=-(N(B_(2)-B_(2))A costheta)/(Deltat)` `=-50(0.35-0.10)xxpi(3xx10^(-2))^(2)xxcos0^(@))/(2xx10^(-3))=17.7V`. |
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| 543. |
A wire loop is rotated in magneitc field. The frequency of change of direction of the induced e.m.f. is.A. Six times per revolutionB. Once per rovolutionC. twice per revolutionD. four times per revolution |
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Answer» Correct Answer - C From the knowledge of theory, the direction of induced e.m.f. will change twice per revolution. |
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| 544. |
A long conducting wire `AH` is moved over a conducting triangular wire `CDE` with a constant velocity `v` in a uniform magnetic field `vec(B)` directed into the plane of the paper. Resistance per unit length of each wire is `rho`. Then A. (a) a constant clockwise induced will flow in the closed loopB. (b) an increasing anticlockwise induced current will flow in the closed loopC. (c ) a decreasing anticlockwise induced current will flow in the closed loopD. (d) an constant anticlockwise induced current will flow in the closed loop |
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Answer» Correct Answer - D (d) Magnetic flux in `ox` direction through the coil is increasing. Therefore, induced current will produce magnetic field in `o.` direction. Thus, the current in the loop is anticlockwise. Magnitude of induced current at any instant of time is `I = (e)/(R) = (Bv(FG))/(rho(FG + GD + DF))` When the wire `AH` moves downwards `FG, GD and DF` all increasing in the same ratio. Therfore, `i` is constant. |
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| 545. |
A 2 m long metallic rod rotates with an angular frequency `200" rod "s^(-1)` about on 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 megnetic field of 0.5 T parallel to axis exises everywhere. The emf developed between the centre and the ring isA. 100VB. 200VC. 300VD. 400V |
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Answer» Correct Answer - B The emf developed between the centre and the rings is `epsi=(1)/(2)Bl^(2)omega=(0.5xx2^(2)xx200)/(2)=200V` |
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| 546. |
In the circuit shown in Fig. A conducting wire HE is moved with a constant speed v towards left. The complete circuit is placed in a uniform magnetic field `vec(B)` perpendicular to the plane of circuit inwards. The current in HKDE is A. (a) clockwiseB. (b) anticlockwiseC. ( c) alternatingD. (d) zero |
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Answer» Correct Answer - D (d) Potential difference across capacitor `V = Bvl = constant` Therefore, charge stored in the capacitor is also constant. athus, current through the capacitor is zero. |
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| 547. |
In the circuit diagram shown in Figure, `R=10 Omega, L=l5H, E=20v, i=2A`. This current is decreasing at a rate of `-1.0A//s.` Find `V_(ab)` at this instant. |
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Answer» `PD` across inductor, `V_L= L (di)/(dt)=(5)(-1.0)=-5V` now, `V_a-iR-V_L=E=V_b` `:. V_(ab) =V_a-V_b=E+iR+V_L` `=20+(2)(10)-5=35V` |
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| 548. |
In the previous question, if the direction of `I` is reversed, `V_(B)-V_(A)` will beA. `20V`B. `15V`C. `10V`D. `5V` |
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Answer» Correct Answer - D `V_(A)-5xx1+15-5=V_(B) implies V_(B)-V_(A)=5V` |
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| 549. |
What are Eddy currents ? |
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Answer» Eddy currents (or) Foucault currents: The induced circulating currents produce in a conductór itself due to change in magnetic flux linked with the conductor are called Eddy currents. Due to Eddy currents, the energy is dissipatėd in the form of heat energy. |
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| 550. |
Describe the ways in which Eddy currents are used to advantage. |
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Answer» Eddy currents are used to advantage in (i) Magnetic braking in trains : A strong magnetic field is applied across the metallic drum rotating with the axle of the electric train. Thus large eddy currents are produced in the metallic drum. These currents oppose the motion of the drum and hence the axle of the train whicn ultimately makes the train come to rest. (ii) Induction Motor: Eddy currents are used to rotate the short circuited rotor of an induction motor. Ceiling fans are also induction motors which run on single phase alternating current. (iii) Electromagnetic damping : Certain galvanometers have a fixed core made of non magnetic metallic material. When the coil oscillates, the eddy currents generated in the core oppose the motion and bring the coil to rest quickly. (iv) Induction furnace : Induction furnace can be used to produc high temperatures anu can be utilised to prepare alloys, by melting the constituent metals. A high frequency alternating current is passed through a coil. The eddy currents generated in the metals produce high temperatures sufficient to melt it. (v) Analogue energy meters : Concept of eddy currents is used in energy meters to record the consumption of electricity. Aluminium disc used in these meters get induced due to varying magnetic field. It rotates due to eddy currents produced in it. |
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