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InterviewSolution
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.
| 1. |
There is a conducting ring of radius `R`. Another ring having current `i` and radius `r (r lt lt R)` is kept on the axis of bigger ring such that its center lies on the axis of bigger ring at a distance `x` from the center of bigger ring and its plane is perpendicular to that axis. The mutual inductance of the bigger ring due to the smaller ring isA. `(mu_(0)piR^(2)r^(2))/((R^(2) + x^(2))^(3//2))`B. `(mu_(0)piR^(2)r^(2))/(4(R^(2) + x^(2))^(3//2))`C. `(mu_(0)piR^(2)r^(2))/(16(R^(2) + x^(2))^(3//2))`D. `(mu_(0)piR^(2)r^(2))/(2(R^(2) + x^(2))^(3//2))` |
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Answer» Correct Answer - D Let current `i` flows in the bigger ring, then the magnetic field on its axis `B = (mu_(0)iR^(2))/(2(R^(2) + x^(2))^(3//2))` Flux linked with the smaller ring : `phi = Bpi r^(2)` `phi = (mu_(0) iR^(2))/(2(R^(2) + x^(2))^(3//2)) pir^(2) = Mi` `:. M = (mu_(0)piR^(2)r^(2))/(2(R^(2) + x^(2))^(3//2))` |
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| 2. |
The capacitor of an oscillatory circuit of frequency `10000 Hz` is enclosed in a container. When the container is evacuated, the frequency changes by`50 Hz`, the delectric constant of the gas isA. `1.1`B. `1.01`C. `1.001`D. `1.0001` |
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Answer» Correct Answer - B `f = (1)/(2pi sqrt(LC)), f + 50 = (1)/(2pi sqrt(LC//K))` `rArr K = ((f + 50)/(f))^(2) = (1 + (50)/(f))^(2)` `rArr K = 1 + (100)/(f) = 1 + (100)/(10,000) = 1.01` |
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| 3. |
The total heat produced in resistor `r` in an `RL` circuit when the current in the inductor decreases from `I_(o)` to `0` isA. `LI_(0)^(2)`B. `(1)/(2) LI_(0)^(2)`C. `(3)/(2) LI_(0)^(2)`D. `(1)/(3) LI_(0)^(2)` |
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Answer» Correct Answer - B the power dissipated in the resistor, `P = (dW)/(dt) = I^(2)R` Since the current through resistor varies with time, we must integrate. The total energy produced as heat in the resistor `W = int_(0)^(oo) I^(2) R dt` The current in an `RL` circuit is `I = I_(0)e^(-(R//L)t)` `W = int_(0)^(oo) I_(0)^(2) e^(-(R//L)t) R dt = (I_(0)^(2) R)/(-2R//L) [e^((-2R)/(L)t)]_(0)^(oo) = (1)/(2)` We can integrate by substituting Note thet the total heat produced equals the energy `(1//2)LI_(0)^(2)` originally stored in the conductor. |
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| 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. The magnitude of the induced electric field at point P at a distance r from the centre of the circular region A. is zeroB. decreases as `1//r`C. increases as `r`D. decreases as `1//r^(2)` |
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Answer» Correct Answer - B `E(2pi r) = pia^(2) (dB)/(dt)` for `r gt = a` `rArr E = (a^(2))/(2r) (dB)/(dt) rArr E prop (1)/(r )` |
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| 5. |
Figure shows two circular regions `R_(1)` and `R_(2)` with redii `r_(1) = 21.2 cm` and `r_(2) = 32.3 cm`, respectively. In `R_(1)` there is a uniform magetic field `B_(1) = 48.6 mT` into the page and in `R_(2)` there is a uniform magnetic field `B_(2) = 77.2 mT` out of the page (ignore any fringing of these fields).Both fields are decreasing at the rate `8.50 mT s^(-1)`. Calculate the intergal `oint vec(E). vec(dl)` for each of the three identical paths. |
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Answer» `oint vec(E) . dvec(s) = - (d phi)/(dt) = - A (dB)/(dt)` For loop a: `oint vec(E) . vec(dl) = - (d(vec(B) . vec(A)))/(dt) = (d(BA cos 180^(@)))/(dt)` `= a(dB)/(dt) = pi r_(1)^(2) (-8.5 mT//s)` `= - pi (21.2 xx 10^(-2))^(2) (8.5 xx 10^(-3))` `= - 1.2 xx 10^(-3) V` For loop b: `oint vec(E) .vec(dl) = - (d(vec(B) .vec(A)))/(dt) = A (dB)/(dt)` `= - pi ((32.3)/(100))^(2) xx 8.5 xx 10^(-3) = - 1.2 xx 10^(-3) V` For loop c: `oint vec(E) .vec(dl) = - (d(vec(B) .vec(A)))/(dt) = - (d(vec(B_(1))*vec(A_(1)) + vec(B_(2))* vec(A_(2))))/(dt)` `= - (d(B_(1)A_(1) cos180^(@) + B_(@)A_(2) cos0^(@)))/(dt)` `= + A_(1) (dB_(1))/(dt) - A_(2) (dB_(2))/(dt)` `= - 1.2 xx 10^(-3) - (-2.7 xx 10^(-3))` `= - 1.5 xx 10^(-3) V` |
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| 6. |
Derive an expression for the total magnetic energy stored in two coils with inductances `L_(1)` and `L_(2)` and mutual inductance `M`, when the currents in the coils are `I_(1)` and `I_(2)`, respectively. |
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Answer» When the currents are increaseing in the circuit, we have emfs `E_(1) = - L_(1) (dI_(1))/(dt) +- M (dI_(2))/(dt)`,`E_(2) = - L_(2) (dI_(2))/(dt) +- M(dI_(1))/(dt)` where `+-` sign appears conistently in both equations and depends on the geometry of the coils and the sense of current. Work done in pushing charges `dq_(1)` and `dq_(2)` through each circuit. respectively, is `dW = - E_(1) dq - E_(2) dq_(2)` `= L_(1) (dI_(1))/(dt) dq_(1) bar+ M (dI_(2))/(dt)dq_(1) + L_(2) (dI_(2))/(dt) dq_(2) bar+ M (dI_(1))/(dt) dq_(2)` `I_(1) = (dq_(1))/(dt),dq_(1) = I_(1)dt, I_(2) = (dq_(2))/(dt),dq_(2) = I_(2) dt` `dW = L_(1)I_(1)dI_(1) + L_(2)I_(2)dI_(2) bar+(MI_(1) dI_(2) + MI_(2) dI_(1))` `= L_(1)I_(1)dI_(1) + L_(2)I_(2)dI_(2) bar+Md(I_(1)I_(2))` On integating the above expression from `0` to final current, we have `U = int dW = L_(1) int_(0)^(I_(1)) I_(1) dI_(1) + L_(2) int_(0)^(I_(2)) I_(2) dI_(2) bar+ M int_(0)^((I_(1)I_(2))) d(I_(1)I_(2))` `= (1)/(2) L_(1)L_(1)^(2) + (1)/(2) L_(2)I_(2)^(2) bar+MI_(1)I_(2)` |
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| 7. |
Figure shows five lettered regions in which a uniform magnetic field extends directly either out of the page (as in region `a`) or into he page. The field is increasing in magnitude at the same steady rate in all five regions, the regions are identical in area. Also shows are four numbered paths along which `oint vec(E). vec(dl)` has the magnitudes given below in terms of a quantity mag. Determine whether the magnetic fields in regions `b` through `e` are directed into ot out of the page. ? `|{:("Path",1,2,3,4),(oint vec(E).vec(dl),mag,2(mag),3(mag),0):}|` |
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Answer» `oint_(1) vec(E)* vec(dl) = mag` `oint_(3) vec(E)* vec(dl) =3(mag)` It means field in `b` and `c` should be out of the page. `oint_(4) vec(E)* vec(dl) = 0` It means field in `e` will be opposite to that in c,i.,e., into the page. Similarly field into `d` is also into the page. |
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| 8. |
A magentic field directed into the page changes with time according to `B = (0.0300t^(2) + 1.4.40)T`, where `t` is in seconds. The field has a circular cross section of radius `R = 2.50 cm`. What are the magitude and direction of the electric field at point `P_(1)` when `t = 3.00 s` and `r_(1) = 0.0200 m`? |
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Answer» `(dB)/(dt) = 0.060 t` `E_(atP_(1)) = (r )/(2) (dB)/(dt) = (0.020)/(2) xx 0.060 = 6 xx 10^(-4) V//m` Here magentic field is increasing with time hence electric field will be in anticlockwise direction. |
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| 9. |
If `epsilon_0` and `mu_o` are, respectively, the electric permittivity and magnetic permeability of free space, `epsilon` and `mu` the corresponding quantities in a medium, the index of refraction of the medium in terms of the above parameters is…. |
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Answer» We know that the velocity of light in vacuum `c = (1)/(sqrtmu_(0)epsilon_(0))` and the velocity of light in a medium `nu = (1)/(sqrt(mu epsilon))*` Also, the refractive index `n = ("Velocity of light in vacuum")/("Velocity of light in medium") = (c )/(nu)` `= (1//sqrt(mu_(0)epsilon_(0)))/(1//sqrt(mu epsilon)) = (sqrt(mu epsilon))/(sqrt(mu_(0)epsilon_(0)))` |
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| 10. |
In an `LC` circuit as shows in Fig. the switch is closed at `t = 0`.`Q_(max) = 100 mu C`, `L = 40 mH`, `C = 100 mu F`. a. Determine the equation for instantaneous change on the capacitor. b. Determine the equation for instantaneous current in the circuit. |
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Answer» a. `q = Q_(max) cos omega t`, Here `Q_(max) = 100 mu C`, `omega = (1)/(sqrt(LC)) =(1)/(sqrt((40 xx 10^(-3))(100 xx 10^(-6)))) = 500 rads^(-1)` `q = 100 cos (500t) mu C` b. `i = (-dq)/(dt) = (100) (500) sin (500t) mu A` or `i = 50000 sin (500t) mu A` |
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| 11. |
The capacitance in an oscilatory `LC` circuit is increased by `1%`. The charge in inductance required to restore its frequency of oscillation is toA. decrease it by `0.5%`B. increase it by `1%`C. decrease it by `1%`D. decrease it by `2%` |
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Answer» Correct Answer - C `omega_(r) =` constant `rArr LC =` constant `rArr L dc + CdL = 0 rArr (dI)/(L) = - (dC)/(C ) = - 1%` |
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| 12. |
In problem `22`, if a capacitor of capacitance `C` is also connected in the larger loop as shows in Fig . , find the charge on the capactior as a function of time. |
| Answer» `q = C varepsilon(1 - e^(-t//RC))` where `varepsilon = (mu_(0)pia^(2))/(b)*` | |
| 13. |
In the given (Fig.) all the symbols have their usual meanings. At `t = 0`, key `K` is closed. Now answer the following questions. At `t = 0`, the euivalent resistance between `A` and `B` isA. `R_(1) + R_(2) + R_(3)`B. `R_(1) + R_(2)`C. `R_(1) + R_(3)`D. indeterminate |
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Answer» Correct Answer - B At `t = 0`, capacitor will behave like a short circuit and the inductor as open circuit but as `t rarr oo`, the nature is just opposite. |
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| 14. |
In the following electrical network at `t = lt 0` Fig key is placed on (1) till the capacitor got fully charged. Key is placed on (2) at `t = 0`. Time when the energy in both the same for the time is A. `(pi)/(4) sqrt(LC)`B. `(3pi)/(4) sqrt(LC)`C. `(pi)/(3) sqrt(LC)`D. `(2pi)/(3) sqrt(LC)` |
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Answer» Correct Answer - A When energy on both is same, means energy on capacitor is half of its maximum energy. `(q^(2))/(2C) = (1)/(2) (Q^(2))/(2C) rArr q = (Q)/(sqrt(2))` `rArr Q cos omega t = (Q)/(sqrt(2)) rArr cos omega t = (1)/(sqrt(2))` `rArr omegat = (pi)/(4) rArr t = (pi)/(4 omega) = (pi)/(4) sqrt(LC)` |
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| 15. |
The network shows in Fig is part of a complate circuit. If at a certain instant the current `(I)` is `5 A`, and is decreasing at a rate of `10^(5) A s^(-1)` then `V_(B) - V_(A) = - V`. |
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Answer» `V_(A) - IR + E - L (dI)/(dt) = V_(B)` `rArr V_(A) - 5 xx 1 + 15 - 5 xx 10^(-3)(-10^(-3)) = V_(B)` `rArr V_(B) - V_(A) = 15V` |
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| 16. |
A field line is shows in Fig. This field cannot represent A. Magnetic fieldB. Electrostatic fieldC. Induced electric fieldD. Gravitational field |
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Answer» Correct Answer - B::D Electrostatic and gravitational field do not make closed loop. |
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| 17. |
Which of the two curves shows has lesser time constant. |
| Answer» Curve 1, because in the same time, current in 1 is more than 2. | |
| 18. |
A closed loop of cross-sectional area `10^(-2) m^(2)` which has inductance `L = 10 mH` and negligible resistance is placed ti time- varying magnetic field. Figure shows the variation of `B` with time for the intervel `4 s`. The field is perpendicular to the plane of the loop (given at `t = 0`, `B = 0`, `I = 0)`. The value of the maximum current induced in the loop is A. `0.1 mA`B. `10 mA`C. `100 mA`D. Date insufficient |
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Answer» Correct Answer - C Induced e.m.f. `e = L(dI)/(dt) = A (dB)/(dt)` `rArr int_(0)^(1) dI = int_(0)^(B) (A)/(L) dB rArr I = (A)/(L) B` `rArr I_(max) = (A)/(L) B_(max) = (10^(-2))/(10 xx 10^(-3)) xx 0.1 = 0.1 A = 100 mA` |
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| 19. |
In the circuit shows in Fig. the battery has negligible internal resistance. Show that the current in the circuit through the battery rises instanlty to its steady state value `E//R` when the switch is closed, provided that resistance `R` is `sqrt(L//C)`. |
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Answer» Let the currents through inductive branch and capacitive branch be `I_(L)` and `I_(C)`, respectively. Then the current through the battery, from `KCL`, is `I = I_(L) + I_(C)`. Since the battery is connected in parallel to the `RL` and `RC` branches of the circuit, the current in the `RL` branch is unaffected by the presence of the `RC` branch, so `I_(L) = (E)/(R) [1 -e^(-(R//L)t)]`and `I_(C) = (E)/(R) e^(-(t//RC))` Hence, the current through battery is `I = (E)/(R) [1 -e^(-(R//L)t) + e^(-(t//RC))]` If the current has to reach its final value `E//R` instananeously, the exponitial terms must cancel out, i.e., `e^(-(t//tauL)) = e^(-(t//tauC))`, which is possible if `tau_(L) = tau_(C)`. `L//R = RC,R = sqrt((L//C))` For at `t gt 0`, this is the desired result. |
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| 20. |
A circuit consists of a coil, a switch and a battery, all in series. The interval resistance of the battery is negligible compared with that of the coil. The switch is originally open. It is thrown closed, after a time interval `Deltat`. The current in the circuit reaches `80.0%` of its final value. The switch then remains closed for a time interval much longer than `Deltat`. The wires connected to the terminals of the battery are then short circuited with another wire and removed from the battery, so that the current is uninterrupted. At the moment `2Deltat` after the coil is short-circuited, the current in the coil is what percentage of its maximum value?A. `20.0%`B. `8.0%`C. `4.0%`D. `10.0%` |
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Answer» Correct Answer - A |
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| 21. |
A solenoid of radius 2.50cm has 400turns and a length of 20.0cm. The current in the coil is changing with time such that to produce an emf of 75.0 mV. the inductance (L) of this solenoid isA. 4mHB. 5mHC. 3mHD. 2mH |
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Answer» Correct Answer - D |
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| 22. |
In the circuit shows Fig the coil has inductance and resistance. When `X` is joined to `Y`, the time constant is `t` 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`: A. the total heat produed in the coil is `p tau`B. The total heat produced in the coil is `(1)/(2) P tau`C. The total heat produced in the coil is `2 P tau`D. The data given are not sufficient to reach a conlusion |
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Answer» Correct Answer - B In steady state: `P = I^(2)R` Energy in inductor `= (1)/(2) LI^(2)` After connecting `x` and `z`, the whole energy of inductor will go into heat so heat produced : `H = (1)/(2) lI^(2) = (1)/(2) L (P)/(R ) = (1)/(2) P tau` |
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| 23. |
The time constant of an inductance coil is `2 xx 10^(-3) s`. When a `90-Omega` resistance is joined in series, the same constant becomes `0.5 xx 10^(-3) s`. The inductance and resistance of the coil areA. `30 mH`, `30 Omega`B. `60 mH`, `30 Omega`C. `30 mH`, `60 Omega`D. `60 mH`, `60 Omega` |
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Answer» Correct Answer - B `(L)/(R ) = 2 xx 10^(-3)` (i) `(L)/(R + 90) = 0.5 xx 10^(-5)` From (i) and (ii), on solving, we get `L = 60 mH` and `R = 30 Omega` |
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| 24. |
A coil of inductance `0.20 H` is connected in series with a switch and a cell of emf `1.6 V`. The total resistance of the circuit is `4.0 Omega`. What is the initial rate of growth of the current when the switch is closed?A. `0.050 A s^(-1)`B. `0.40 A s^(-1)`C. `0.13 A s^(-1)`D. `8.0 A s^(-1)` |
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Answer» Correct Answer - D `V = RI + L(dI)/(dt)`, at `t =0, I = 0`, thus we have `(dI)/(dt) = (V)/(L) = (1.6)/(0.2) = 8 A//s` |
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| 25. |
Figure shows a circuit having a coil of resistance `R = 2.5 Omega` and inductance `L` connected to a conducting rod of radius `10 cm` with its center at `P`. Assume that friction and gravity are absent and a constant uniform magnatic field of `5 T` exists as shown in figure. At `t = 0`, the circuit is switched on and simultaneously a time-varying external torque is applied on the rod so that it rotates about `P` with a constant angular velocity `40 rad s^(-1)`. Find the magnitude of this torque (in `mNm`) when current reaches half of its maximum value. Neglect the self inductance of the loop formed by the circuit. |
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Answer» Correct Answer - `(5)` Induced `EMF = (1)/(2) B omega l^(2)` Maximum current: `i_(0) = (B omegal^(2))/(2R)` Torque about the hinge `P` is `tau = underset(0) overset(l) int i(dx)Bx rArr tau = (1)/(2) iBl^(2)` Putting `i= i_(0)//2`, we get: `tau = (B^(2)Omega l^(4))/(8R) = 5mNm` |
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| 26. |
The network shown in the figure is a part of complete circuit. What is the potential difference `V_B-V_A` when the current `I` is `5 A` and is decreasing at a rate of `10^3A//s`? |
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Answer» In accordance with the law of potential distribution, for the given network, `V_(A) - IR + E - (dI)/(dt) L = V_(B)` And here `I` is decreasing (i.e., `dI//dt` is negative). `V_(B) - V_(A) = - 5 xx 1 + 15 - 5 xx 10^(-3)(-10^(3))` `V_(B) - V_(A) = - 5 + 15 + 5 = 15 V` |
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| 27. |
The inductance `L` of a solenoid of length `l`, whose windings are made of material of density `D` and resistivity `rho`, is (the winding resistance is`R`)A. `(mu_(0))/(4 pil) (Rm)/(rho D)`B. `(mu_(0))/(4piR) (lm)/(rho D)`C. `(mu_(0))/(4pil) (R^(2)m)/(rhoD)`D. `(mu_(0))/(2 pi R) (lm)/(rhoD)` |
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Answer» Correct Answer - A For a solenoid, `L = mu_(0)N^(2) (A)/(l)*` if `x` is the length of the wire and `a` is the area of cross-section, then `R = (rho x)/(a)` and `m = axD` `Rm = (rhox)/(a) axD, x = sqrt((Rm)/(rho D))` Also, `x = 2 pi rN, N = (x)/(2 pir) ( :. L = (mu_(0)N^(2)A)/(l))` `:. L = mu_(0)((x)/(2pir))^(2) (pir^(2))/(l) = (mu_(0))/(4 pil) (Rm)/(rhoD)` |
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| 28. |
Current in a coil of self-inductance `2.0 H` is increasing as `I = 2 sin t^(2)`. The amount of energy spent during the period when the current changes from `0` to `2 A` isA. `1 J`B. `2 J`C. `3 J`D. `4 J` |
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Answer» Correct Answer - D energy spent `= (1)/(2) Ll^(2)` |
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| 29. |
A self-induced wmf in a solenoid of inductance L changes in time as`varepsilon=varepsilon_0e^(-kz).` Assuming the charge is finite.find the total charge that passes a point in the wire of the solenoid.A. `varepsilon_0/(Lk^2)`B. `(2varepsilon_0)/(Lk^2)`C. `varepsilon_0/(2Lk^2)`D. `(3varepsilon_0)/(Lk^2)` |
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Answer» Correct Answer - A |
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| 30. |
A toroid is wound over a circular core. Radius of each turn is `r` and radius of toroid is `R (gt gt r)`. The coefficient of self-inductance of the toroid is given by A. `L = (mu_(0)Nr^(2))/(2R)`B. `L = (mu_(0)Nr)/(2R)`C. `L = (mu_(0)Nr^(2))/(R )`D. `L = (mu_(0)N^(2)r^(2))/(2R)` |
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Answer» Correct Answer - D `L = (phi)/(I),phi = NAB` `N = mu_(0)nI` where `n = (N)/(2 pi R)` `:. phi = Npi r^(2) (mu_(0)(N)/(2 piR)I)` `phi = (mu_(0)N^(2)r^(2)I)/(2R)` `L = (phi)/(I) = (mu_(0)N^(2)r^(2))/(2R)` |
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| 31. |
Two insulated wires are wound on the same hollow cylinder, s as to from two solenoids shering a common air-filled core. Let `l` be the length of the core, A the cross -sectional area of the core, `N_(1)` the number of times the first wire is wound around the core, and `N_(2)` the number of times the second wire is wound around the core. Find the mutual inductance of the two solenoids, neglecting the end effects. |
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Answer» If a current `I_(1)` flows around the first wire then a unfirom axial magnetic field of strength `B_(1) = (mu_(0)N_(1)I_(1))/(l)` is generated in the core. The magnetic field in the region outside the core is of negligible magnitude.The flux linked to a single turn of the second wire is `B_(1)A`. Thus, flux linked to all `N_(2)` turns of the second wire is Now, `phi = N_(2)B_(1)A = (mu_(0)N_(1)N_(2)AI_(1))/(l) = MI_(1)` `:. M = (mu_(0)N_(1)N_(2)A)/(l)` As described pervisously, `M` is a genometric quantity depending on the dimensions of the core and the manner in which the two wires are wound around the core, but not no the actual actual currents flowing through the wires. |
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| 32. |
Figure. shows two concentric coplanar coils with radii `a` and `b(alt ltb)`. A current `I = 2t` flows in the smaller loop. Neglecting self-inductance of the larger loop, a. find the mutual inductance of the two coils, b. find the emf induced in the larger coil, c. if the resistance of the larger loop is `R`, find the current in it as a function of time. |
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Answer» a. To find mutual inductance, it does not matter in which coil we consider current and in which coil flux is calculated (reciprocity theorem).Let current `i` be flowing in the larger coil. Magnetic field at the centre `= (mu_(0)i)/(2b)*` Flux through the smaller coil `= (mu_(0)i)/(2b) pia^(2)` `:. M = (mu_(0))/(2b) pia^(2)` b. |emf induced in larger coil|` = [((di)/(dt))` in smaller coil] `= (mu_(0))/(2b) pia^(2)(2) = (mu_(0)pia^(2))/(b)` c. current in the larger coil `= (mu_(0)pia^(2))/(bR)*` |
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| 33. |
The approximate formula expressing the formula of mutual inductance of two coaxial loops of the same redius `a` when their centers are separated by a distance `l` with `l gt gt a` isA. `(1)/(2) (mu_(0)pia^(4))/(l^(3))`B. `(1)/(2) (mu_(0)a^(4))/(l^(2))`C. `(mu_(0))/(4pi) (pia^(4))/(l^(2))`D. `(mu_(0))/(pi) (a^(4))/(l^(3))` |
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Answer» Correct Answer - A 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 = (mu_(0))/(4pi) (2I pia^(2))/((a^(2) + l^(2))^(3//2))` where `p_(m) = I pia^(2)` = magnetic moment of the loop Flux through the other loop is `phi_(12) = Bpi a^(2)` `= (mu_(0))/(4pi) (2I pia^(2))/((a^(2) + l^(2))^(3//2)) pi a^(2)` or `M = (phi_(12))/(I) = (mu_(0))/(4pi) (2I pia^(4))/((a^(2) + l^(2))^(3//2))` `= (mu_(0) pia^(4))/(2(a^(2) + l^(2))^(3//2))= (mu_(0)pia^(4))/(2l^(3)) (alt lt l)` |
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| 34. |
Switch `S` of the circuit shows in Fig. is closed at `t = 0`. If `e` denotes the induced emf in `L` and `i` the current flowing through the circuit at time `t`, then which of the following graphs correctly represents the variation of `e` with `i`? B. C. D. |
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Answer» Correct Answer - B `e = E - iR` Clearly, the graph is a straight line with negative slope. |
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| 35. |
In the circuit shows in Fig switch `S` is shifed to position `2` from position `1` at `t = 0`, having been on position `1` for a long time. The current in the circuit just after shifting of switch will be (battery and both the inductors are ideal) A. `(4)/(5) (E)/(R )`B. `(5)/(4) (E)/(R )`C. `(5)/(9) (E)/(R )`D. `(E)/(R )` |
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Answer» Correct Answer - C Let `i_(1)` be the current in the circuit before shifting `i_(1) = (E)/(R )` (i) Since the flux associated with the inductors will be same just before and juse after shifting, therefore `i_(1) 5L = i_(2) 9L` `i_(2) = (5)/(9) (E)/(R )` |
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| 36. |
Solve problem `19` if the planes of the coils make an angle `theta` with each other. |
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Answer» If current `i` flows in the larger coil, magnetic field produced at the centre will be perpendicular to the plane of the larger coil. Now, the area vector of the smaller coil which is per pendicular to the plane of the smaller coil will make an angle `theta` with the magnetic field. Thus, flux `vec(B)*vec(A) = (mu_(0)i)/(2a_(2)) pia_(1)^(2) costheta` or `M = (mu_(0)pia_(1)^(2) cos theta_(1))/(2a_(2))` |
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| 37. |
Solve problem `19` if the planes of the coils are perpendicular. |
| Answer» Let a current `i` flows in the coil of radius `a_(1)`. The magnetic field at the centre of this coil will now be parallel to the plane of the smaller coil and hence no flux will pass through it, so `M = 0`. | |
| 38. |
It has been proposed to use large inductors as energy storage devices. a. How much electrical energy is converted to light and thermal energy by a `200 - Omega` light bulb in one day? b. If the amount of energy calculated in part (a) is stored in an inductor in which the current is `80.0 A`, what is the inductance? |
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Answer» a. `U = Pt = (200 W)(24 h//day) xx 3600 s//h) = 1.728 xx 10^(7) J` b. `U = (1)/(2) LI^(2) rArr L = (2U)/(I^(2)) = (2(1.728 xx 10^(7)J))/((80.0 A)^(2) = 5400 H` |
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| 39. |
A capacitor with capacitance `6 xx 10^(-5) F` is charged by connecting it to a `12 -V` battery. The capacitor is disconnected from the battery and connected across an inductor with `L = 1.50 H`. a. What are the angular frequency `omega` of the electrical oscillations and the period of these oscillations (the time for one oscillation)? b. What is the intial charge on the capacitor? c. How much energy is intially stored in the capacitor? d. What is the charge on the capacitor `0.0230` s after the connecting to the inductor is made? Interpret the sign of the your answer. e. At the times given in part (d), what is the current in the inductor? Interpret the sign of your answer. f. At the time given in part (d), how much electrical energy is stored in the capactior and how much is stored in the inductor? |
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Answer» a. `T = (2 pi)/(omega) = 2 pi sqrt(LC) = 2 pi sqrt((1.50 H)(6.00 xx 10^(-5) F)` `= 0.0596 s`, `omega = 105 red//s` b. `Q = CV = (6.00 xx 10^(-5) F)(12.0 V)= 7.20 xx 10^(-4) C` c. `U_(0) = (1)/(2) CV^(2) = (1)/(2) (6.00 xx 10^(-5) F)(12.0 V)^(2) = 4.32 xx 10^(-3) J` d. `t = 0.230 s`, `q = Q cos (omega t)` `= (7.20 xx 10^(-4) C) xx cos((0.0230 _(S))/sqrt((1.50 H)(6.00 xx 10^(-5) F)))` `= - 5.43 xx 10^(-4) C` Negative sign indicates that signs on plates are opposite to those at initials e. `t = 0.0230 s`, `i = (dq)/(dt) = - mu Q sin (omega t)` `i = - (7.20 xx 10^(-4)C)/sqrt((1.50 H)(6.00 xx 10^(-5) H))` `xx sin ((0.0230_(S))/sqrt((1.50 H)(6.00 xx 10^(-5) H))) = - 0.05 A` Negative sign indicates that positive charge flowing away from plate which had positive charge at the given time. f. Capacitor : `U_(C) = (q^(2))/(2C) = ((5.43 xx 10^(-4) C)^(2))/(2(6.00 xx 10^(-5) F)) = 2.46 xx 10^(-3)J` Inductor : `U_(L) = (1)/(2) Li^(2) = (1)/(5) (1.50 H)(0.05 A)^(2)` |
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| 40. |
Solenoid `S_1` has N turns, radius `R_1` and length l. It is so long that its magnetic field is uniform nearly everywhere inside it and is nearly zero outside, Solenoid `S_2` has `N_2` turns, radius `R_2ltR_1`.and the same length as `S_1` It lies inside `S_1` with their axes prallel. (a) Assume `S_1` carries variable current i. Compute the mutual inductance characterizing the emf induced is `S_2`. (b) Now assume `S_2` carries current i. Compute the mutual inductance to which the emf in `S_1` is proportional. (c) State how we results of parts (a) and (b) compare with each other. |
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Answer» Correct Answer - A::B::C |
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| 41. |
A long solenoid of radous `R` has `n` turns of wire per unit length and carries a time-varying current that varies sinusiodally as `I = I_(max) cos omegat`, where `I_(max)` is the maximum current and `omega` I the angular frequancy of the alternating current source (shows in Fig.) The magnitude of the induced electric field inside tha solenoid, a distance `r lt R` from its long central axis isA. `(3 mu_(0)nI_(max)omega)/(2) r sin omega t`B. `(mu_(0)nI_(max)omega)/(2) r cos omega t`C. `mu_(0)nI_(max)omega r sin omega t`D. `(mu_(0)nI_(max)omega)/(2) r sin omega t` |
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Answer» Correct Answer - D `E = (r )/(2) (dB)/(dt) = (r )/(2) mu_(0)n (dI)/(dt) rArr E = - (mu_(0)nr)/(2) I_(max) omega sin omega t` |
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| 42. |
Switch `S` shown in Fig. is closed for `t lt 0` and is opened at `t = 0`. When currents through `L_(1)` and `L_(2)` are equal, their common value is A. `(E)/(R )`B. `(E(L_(2) + L_(2)))/(RL_(1))`C. `(EL_(1))/(R(L_(1) + L_(2)))`D. `(E)/(R ) ((L_(1) + L_(2)))/(L_(2))` |
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Answer» Correct Answer - C Constancy of flux implies that `(E)/(R ) L_(1) = i(L_(1) + L_(2))` i.e., `i = (EL_(1))/(R(L_(1) + L_(2))` |
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| 43. |
In Fig. switch `S` is closed for a long time. At `t = 0`, if it is opened, then A. total heat produced in resistor `R` after opening the switch is `(1)/(2) (Lepsilon)/(R^(2))`B. total heat produced in resistor `R_(1)` after opening the switch is `(1)/(2) (Lepsilon^(2))/(R^(2)) ((R_(1))/(R_(1) + R_(2)))`C. heat produced in resistor `R_(1)` after opening the switch is `(1)/(2) (R_(2)L epsilon^(2))/((R_(1) + R_(2)) R^(2))`D. no heat will be produced in `R_(1)` |
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Answer» Correct Answer - C Just before opening the switch, the current in the inductor is `(E)/(R )*` Energy stored in it `= (1)/(2) L((epsilon)/(R ))^(2)` This energy will dissipate in resistance `R_(1)` and `R_(2)` in the ratio `(1)/(R_(1))` and `(1)/(R_(2))*` |
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| 44. |
In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch `(S_1)`. Also when `(S_1)` is opend and `(S_2)` is closed the capacitor is connected in series with inductor (L). When the capacitor gets charged compleely, (=`(S_1)` is opened and `(S_2)` is closed, Then,A. at `t = 0`, energy stored in the circuit is purely in the from of magnetic energy.B. at any time `t gt 0` current in the circuit is in the same direction.C. at `t gt 0`, there is no excharge of energy between the inductor and capacitor.D. at any time `t gt 0`, maximum instantaneous current in the circuit may be `V sqrt((C )/(L))*` |
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Answer» Correct Answer - D From conservation of energy, `(1)/(2)LI_(max)^(2) = (1)/(2)CV^(2) rArr I_(max) V sqrt((C )/(L))` |
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| 45. |
In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch `(S_1)`. Also when `(S_1)` is opend and `(S_2)` is closed the capacitor is connected in series with inductor (L). At the start, the capicitor was uncharged. when switch `(S_1)` is closed and `(S_2)` is kept open, the time constant of this circuit is `tau`. which of the following is correctA. After time interval `tau`, charge on the capacitor is `CV//2`B. After time interval `2 tau`, charge on the capacitor of `CV(1 - e^(-2)]`C. The work done by the voltage source will be half of the heat disspated when the capacitor is fully chargedD. After time interval `2 tau`, charge on the capacitor is `CV(1 - e^(-1)]` |
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Answer» Correct Answer - B Charge on capacitor at time `t` is `q = q_(0) (1 - e^(-t//tau))` Here, `q_(0) = CV` and `t = 2 tau` `q = CV(1 - e^(-2tau//tau)) = CV (1 - e^(-2))` |
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| 46. |
In the circuit shows in Fig. switch `k_(2)` is open and switch `k_(1)` is closed at `t = 0`. At time `t = t_(0)`, switch `k_(1)` is opened and switch `k_(2)` is simultaneosuly closed. The variation of inductor current with time is B. C. D. |
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Answer» Correct Answer - A For `t lt t_(0), E = L(di)/(dt) rArr I = (E)/(L) t` For `t gt t_(0), L(di)/(dt) = 0 rArr i =` constant |
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| 47. |
In the given circuit, initially switch `S_(1)` is closed, and `S_(2)` and `S_(3)` are open. After charging of capacitor, at `t = 0`, `S_(1)` is opened and `S_(2)` and `S_(3)` are closed. If the relation between inductance, capacitance and resistance is `L = 4 CR^(2)`, tehn find the time (in s) after which current passing through capacitor and inductor will be same (given `R = In2 m Omega,L = 2 mH`) |
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Answer» Correct Answer - `(1)` After charging, charge on capacitor `= C epsilon` Now, at `t = 0` two circuits are formed 1. Discharging of capacitor `q = C epsilon e^(-t//tau_(C)) = Cepsilon e^(-t//RC)` `i_(1) = - (dq)/(dt) = (epsilon)/(2R) e^(-t//RC)` 2. Growth of current in `L - R` circuit. `i_(2) = (epsilon)/(2R) [1 - e^(-t//tau_(L))]` Now, `i_(1) = i_(2)` `(epsilon)./(2R) e^(-t//tau_(C)) = (epsilon)/(2R) [1 - e^(-t//tau_(L))]` ..(i) Given `l = 4CR^(2)` `(L)/(2R) = 2RC = (1)/(In 2) rArr tau_(C) = tau_(L) = (1)/(In 2)` Solve to get `t = 1 s` |
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| 48. |
Which of the field patterns given below is valid for electric field as well as for magnetic field?A. B. C. D. |
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Answer» Correct Answer - C True foe induced electric field and magnetic field. |
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| 49. |
Find the current passing through battery immediately after key `(K)` is closed. It is given that initially all the capacitors are uncharged. (Given that `R=6Omega` and `C=4muF)` A. `1 A`B. `5 A`C. `3 A`D. `2 A` |
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Answer» Correct Answer - A `R_(eq) = (5R)/(6) rArr I = (6E)/(5R) = 1 A` |
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| 50. |
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(altltR) having two turns is placed with its centre at `=sqrt(3)R` along the axis of hte circular wire loop, as shown in figure. The plane of the square loop makes an angle of `45^(@)` with respect to the z-axis. If the mutual inductance between the loops is given bu `(mu_(0)a^(2))/(2^(p//2)R)`, then the value of p is |
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Answer» Correct Answer - `7.` `B = (mu_(0)iR^(2))/(2(R^(2) xx X^(2))^(3//2))` `B = (mu_(0)iR^(2))/(2(R^(2) xx 3R^(2))^(3//2)) = (mu_(0)iR^(2))/(2(4R^(2))^(3//2)) = (mu_(0)iR^(2))/(2.2^(3)R) = (mu_(0)i)/(16R)` `phi = 2 (mu_(0)i)/(16R) a^(2) (1)/(sqrt(2)) = (mu_(0)ia^(2))/(8sqrt(2)R)` `M = (phi)/(i)` `M = (mu_(0)a^(2))/(2^(7//2)R) = (mu_(0)a^(2))/(2^(p//2)R)` `p = 7` |
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