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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.
| 101. |
A square metal wire loop of side 10 cm and resistance 1 ohm is moved with a constant velocity `(v_0)` in a uniform magnetic field of induction `B=2 weber//m^(2)` as shown in the figure. The magnetic field lines are perpendicular to the plane to the loop (directed into the paper). The loop is connected to a network of resistors each of value 3 ohms. The resistances of hte lead wire OS and PQ are negligible. What should be the speed of the loop so as to have a steady current of 1 milliampere in the loop? Given the direction of current in the loop. |
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Answer» Network `AQCS` is balanced Wheatone bridge, no current in branch `AC` `R_(eq)=(6xx6)/(6+6)=3Omega` Resistance of loop `=1Omega` `R_(total)=1+R_(eq)=1+3=4Omega` Induced emf, `e=Bv_(0)l` Induced current `I=(e)/(R_(total))=(Bv_(0)l)/(4)` `10^(-3)=(2v_(0)xx0.1)/(4)` implies `v_(0)=2xx10^(-2)m.//sec=2cm//sec` Flux is decreasing, induced current `I` will be clockwise. |
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| 102. |
Find the total heat produced in the loop of the previous problem during the interval `0` to `5 s` |
| Answer» Correct Answer - B | |
| 103. |
In the previous problem, fin dthe total heat produced in the loop. |
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Answer» The current will exist for time `t_(0)=(2l)/(v)` `=((Bvt)/(R))^(2)R((2l)/(v))` `=(2B^(2)vt^(3))/(R)` |
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| 104. |
Find the total heat produced in the loop of the previous problem during the interval 0 to 30 s if the resistance of the loop is `4.5 Omega`.A. `2.5xx10^(4)J`B. `2.0xx10^(4)J`C. `3.0xx10^(4)J`D. `3.5xx10^(4)J` |
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Answer» Correct Answer - B The current exists in the loop for time-interval `t=0` to `t=5s` and `t=20` to `t=25s` i.e., for time `t=10sec` `H=(e^(2))/(R)t=(3xx10^(-4))xx10/(4.5xx10^(-3))=2xx10^(-4)J` |
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| 105. |
Two infinite long straight parallel wires A and B are separted by 0.1m distance and carry equal currents in opposite directions. A square loop of wire C of side 0.1m lies in the plane of A and B. The loop of wire C is kept parallel to both A and B at a distance of 0.1m from the nearest wire. Calculate the EMF induced in the loop C while the currents in A and B are increasing at the rate of `10^(3) A//s`. Also indicate the direction of current in the loop C. |
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Answer» Correct Answer - `2xx10^(-5) In ((4)/(3))` volt , clockwise |
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| 106. |
An AC circuit consists of a resistor of `5Omega` and inductor of 10 mH connected in series with 50 volt, 50 Hz supply. The capacitance that should be connected in series with the circuit to obtain maximum current isA. `1014muF`B. `10.14muF`C. `1.014muF`D. `101.4muF` |
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Answer» Correct Answer - A `f=(1)/(2pisqrt(LC))" "therefore f^(2)=(1)/(4pi^(2)LC)` `therefore C=(1)/(4pi^(2)Lf^(2))` `=(1)/(4xx9.87xx10xx10^(-3)xx2500)=1014muF` |
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| 107. |
An AC circuit consists of a resistor of `5Omega` and inductor of 10 mH connected in series with 50 volt, 50 Hz supply. The capacitance that should be connected in series with the circuit to obtain maximum current isA. 1 mHB. 1.5 mHC. 2.5 mHD. 2 mH |
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Answer» Correct Answer - A `f=(1)/(2pisqrt(LC))` `f^(2)=(1)/(4pi^(2)LC)` `C=(1)/(4pi^(2)Lf^(2))` `=(1)/(4xx9.87xx10xx10^(-3)xx2500)` `=1xx10^(-3)F` =1mF |
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| 108. |
Reactance of a capacitor of capacitance `CmuF` for ac frequency `400/pi Hz` is `25 Omega` . The value C isA. `50muF`B. `25muF`C. `100muF`D. `75muF` |
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Answer» Correct Answer - A `X_(C)=(1)/(2pifC)` `therefore C=(1)/(2pifX_(C))=(1)/(2pixx(400)/(pi)xx25)=50muF` |
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| 109. |
In the given L-R circuit, which of the following gives correct relation between V,`V_(R) and V`? A. `V lt V_(R)+V_(L)`B. `V gt V_(R)+V_(L)``C. `V=V_(R)+V_(L)`D. none of these |
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Answer» Correct Answer - A `e=sqrt(e_(R)^(2)+e_(L)^(2))` `V=sqrt(V_(R)^(2)+V_(L)^(2))` In series circuit due to voltage magnification voltage across L and voltage across R is greater than applied voltage. `therefore V lt V_(R) +V_(L)`. |
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| 110. |
When the primary current in the spark-coil of a car changes from 4A to zero in `10muS`, an emf of 40,000 V is induced in the secondary. The mutual inductance between the primary and the secondary windings of the spark-coil will be -A. 1HB. 0.1HC. 10 HD. zero |
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Answer» Correct Answer - B `e=(M Deltai)/(Deltat)=4000 =Mxx4/(10xx10^(-6))` `M=0.1 H` |
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| 111. |
The displacement current was first postulated byA. CoulombB. FaradayC. MaxwellD. Van-de Graff |
| Answer» Correct Answer - C | |
| 112. |
The self inductance of a solenoid of length L, area of cross-section A and having N turns is-A. `(mu_(0)N^(2)A)/L`B. `(mu_(0)NA)/L`C. `mu_(0)N^(2)LA`D. `mu_(0)NAL` |
| Answer» Correct Answer - A | |
| 113. |
The self inductance of a solenoid of length L, area of cross-section A and having N turns is-A. `(mu_(0)N^(2)S)/(L)`B. `(mu_(0)NS)/(L)`C. `mu_(0)N^(2)LS`D. `mu_(0)NLS` |
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Answer» Correct Answer - A Self-inductance of solenoid is given by `(mu_(0)N^(2)S)/(L)` |
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| 114. |
A long solenoid of `N` turns has a self-induced `L` and area of cross-section A. When a current `i` flows through the solenoid, magnetic field inside it has magnitude `B` . The current `i` is equal toA. `(BAN)/(L)`B. `BANL`C. `(BN)/(AL)`D. `(B)/(ANL)` |
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Answer» Correct Answer - A `L=(mu_(0)N^(2)A)/(l) implies mu_(0)=(Ll)/(N^(2)A)i` `i=(BNA)/(L)` |
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| 115. |
In a cylindrical region uniform magnetic field which is perpendicular to the plane of the figure is in increasing with time and a conducting rod `PQ` is placed in the region.If `C` is the centre of the circle then A. `P` will be at higher potential than `Q`.B. `Q` will be at higher potential than `P`.C. Both `P` and `Q` will be equipotental.D. no `emf` will be developed across rod as it is not crossing /cutting any line of force. |
| Answer» Correct Answer - B | |
| 116. |
State whether the following statement are true of false giving reason in brif : (a) The dimension of `(h//e)` is the same as that of magnetic flux `phi`. (b) The dimensions of electric and magnetic flux are same. (c ) A coil or a metal wire is kept stationary in a non-uniform magnetic field. An e.m.f. is induced in the coil. (d) An e.m.f. can be induced between the two ends of a straingth copper wire when it is moved through a magnetic field. |
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Answer» (a) True. `(h)/(e) = ("joule - sec")/("coulomb") = ("joule")/("ampere")` = magnetic flux (b) False, for reason, see text. (c ) False, As magnetic flux does not change with time. (d) True, as `e = B upsilon l sin theta`. |
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| 117. |
Three identical coils A, B and C are placed with their planes parallel to one another. Fig Coils B and C carry currents as shwon. Coils B and C are fixed. The coil A is moved towards B with uniform speed. Is three any induced current in B |
| Answer» As coils B and C fixed, there is no induced current in B due to C. However, when A is moved towards B, an e.m.f. is induced in B, which would oppose relative motion of A towards B. As current in A is clockwise, induced current in B must be anticlockwise. | |
| 118. |
Three identical coils `A,B` and `C` carrying currents are placed co-axially with their planes parallel to one another. `A` and `C` carry current as shown in figure `B` is kept fixed while `A` and `C` both are moved towards `B` with the same speed. Initially, `B` is equally separated from `A` and `C`. The direction of the induced current in the coil `B` is A. (a) same as that in coil `A`B. (b) same as that in coil `B`C. ( c) zeroD. (d) none of these |
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Answer» Correct Answer - C (c ) Because `A and C` are at equal distance from `B`, and their flux across `B` is in opposite direction, so at any time flux in `B` will be zero. Hence no emf is induced. |
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| 119. |
In co-axial concentric coils of radius `r_(1)` and `r_(2)` such that `r_(1) lt lt r_(2)` Fig. find direction of induced current when K is (i) pressed (ii) released. |
| Answer» (i) When key K is pressed, current in outer coil increases from 0 to maximum. Current induced in inner coil opposes the growth of current in outer coil, by flowing in the opposite direction, i.e. in anticlock wise direciton, as shown in Fig. (ii) When key K is released, current in outer coil decreases from maximum to zero. Current induced in inner coil opposes the decay of current in outer coil, by flowing in the same direction, i.e. in clockwise direction as shown in Fig. | |
| 120. |
Two coils – 1 and 2 – are mounted co axially as shown in the figure. The resistance of the two coils are `R_(1)` and `R_(2)` and their self inductances are `L_(1)` and `L_(2)` respectively. Switch S is closed at time t = 0 to connect the coil 1 to an ideal cell of emf V. It is observed that by the time current reaches its steady value in coil 1, the quantity of charge that flows in coil 2 is `Q_(0)`. Calculate the mutual inductance (M) between the two coils. |
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Answer» Correct Answer - `(R_(1)R_(2)Q_(0))/(V)` |
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| 121. |
Two coils A and B are mounted co-axially some distance apart. Coil A is given a current that changes sinusoidally with time. A current gets induced in coil B. How does the magnitude of current in coil B change if a metal plate is placed between the two coils. |
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Answer» Correct Answer - Current decreases |
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| 122. |
Two circular coils, one of smaller radius `r_(1)` and the other of very large radius `r_(2)` are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement. |
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Answer» Let a current `I_(2)` flow through the outer circular coil. The field at the centre of the coil is `B_(2) = µ_(0)I_(2) // 2r_(2)`. Since the other co-axially placed coil has a very small radius, `B_(2)` may be considered constant over its cross-sectional area. Hence, `Phi_(1)=pir_(1)^(2)B_(2)` `=(mu_(0)pir_(1)^(2))/(2r_(2))I_(2)` `M_(12)I_(2)` Thus, `M_(12)=(mu_(0)pir_(1)^(2))/(2r_(2))` From Eq. (6.14) `M_(12)=M_(21)=(mu_(0)pir_(1)^(2))/(2r_(2))` Note that we calculated `M_(12)` from an approximate value of `Phi_(1)`, assuming the magnetic field `B_(2)` to be uniform over the area `pir_(1)^(2)`. However, we can accept this value because `r_(1) gtgt r_(2)`. |
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| 123. |
Two concentric circular coil,s one of small radius `r_(1)` and the other of large radius `r_(2)`, such that `r_(1) lt lt r_(2)`, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement. |
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Answer» Let a current `I_(2)` flow through the outer circuler coil. The field at the centre of the coil is `B_(2)=mu_(0)I_(2)//2r_(2)`. Since the other co-axially placed coil has a very small radius `B_2` may be considered constant over its cross-sectional area. Hence, `phi_(2)=pir_1^(2)B_(2)` `=(mu_(0)pir_(1)^(2))/(2r_(2))I_(2)` `=M_(12)I_(2)` Thus, `M_(12)==(mu_(0)pir_(1)^(2))/(2r_(2))` From equation `M_(12)=M_21=(mu_(0)pir_(1)^(2))/(2r_(2))` Note that we calculate `M_12` from an approximate value of `phi_(1)` assuming the magnetic field `B_(2)` to be uniform over the area `pi r_1^(2)`. However, we can accept this value because `r_(1) lt lt r(2)`. |
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| 124. |
STATEMENT - 1 : When the current decreases in a conducting loop placed co-axially with a similar loop without current, they attract each other. and STATEMENT - 2 : As the current in one loop decreases, an induced current starts flowing, in same sense, in the other loop.A. Statement-1 is True, Statement-2, is True, Statement-2 is a correct explanation for Statement-2B. Statement-1 is True, Statement-2, is True, Statement-2 is NOT a correct explanation for Statement-2C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - A | |
| 125. |
A conducting circular loop is placed in a uniform magnetic field `0.04T` with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at `2mm//sec` . The induced emf in the loop when the radius is `2cm` isA. `3.2pimuV`B. `4.8pimuV`C. `0.8pimuV`D. `1.6pimuV` |
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Answer» Correct Answer - A `phi=BA=Bpir^(2)` `e=(dphi)/(dt)=Bpi.2r(dr)/(dt)=0.04pixx2xx2xx10^(-2)xx2xx10^(-3)` `=3.2pimuV` |
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| 126. |
A conducting circular loop is placed in a uniform magnetic field, `B=0.025T` with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of `1 mms^(-1)`. The induced emf when the radius is `2 cm` isA. `2pi mu V`B. `pimuV`C. `(pi)/2 mu V`D. `2muV` |
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Answer» Correct Answer - B Magnetic flux `phi=B.A =B pi r^(2)` `EMF |e| =(dphi)/(dt)=Bpi2r (dr)/(dt) =pimuV` |
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| 127. |
The electric mains in a house are marked 220V-50Hz. Write down the equation for instantaneous voltage. |
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Answer» `E_(v) = 200 V, v = 50 Hz E = ?` `E_(0) = sqrt2 E_(v) = 1.414 xx 220 = 311 V` `E = E_(0) sin omega t = 311 sin 2 pi xx 50 t` `= 311 sin 100 pi t` |
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| 128. |
A resistor of `200 Omega`and a capacitor of `15.0muF` are connected in series to a `220V`,` 50Hz` source. (a) Calculate the current in the circuit . (b) Calcutalte the voltage (rms) across the resistor and the inductor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox. |
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Answer» Here, `R = 200 ohm`, `C = 15.0 mu F = 15 xx 10^(-6) F` `E_(v) = 220 V, v = 50 Hz, I_(v) = ?` `V_(R ) = ?, V_(C ) = ?` Now `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)` `= (1)/(2 xx 3.14 xx 50 xx 15 xx 10^(-6)) = 212.3 Omega` (a) Impedance of the circuit, `Z = sqrt(R^(2) + X_(C)^(2))` `= sqrt(2000^(2) + (212.3)^(2))` `= 291.7 ohm` `:.` Current in the circuit, `I_(v) = (E_(v))/(Z) = (220)/(291.7) = 0.75 A` (b) `V_(R ) = I_(v) xx R = 0.75 xx 200 = 150.8 V` `V_(C ) - I_(v) X_(C ) = 0.75 xx 212.3 = 159.2 V` `V_(R ) + V_(C ) = 150.8 + 159.2 = 310 V`, which is more than the source voltage of 220 V. This paradox is resolved by the fact that the two voltages are not in same phase. Therefore, they connot be added like ordinary number. As `V_(R )` and `V_(C )` are out of phase by `90^(@)`, therefore, `V_(RC) = sqrt(V_(R )^(2) + V_(C )^(2)) = sqrt((150.8)^(2) + (159.2)^(2))` `= 220 V`, the source voltage. |
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| 129. |
In L-R circit, the A.C. source has voltage `220V`. If potential difference across inductor is `176V`, the potential difference across the resistor (in Volts) is `Kxx33`. Find the value of `K`A. 13.2 VB. 12 VC. 132 VD. 1.32 V |
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Answer» Correct Answer - C `e=sqrt(e_(R)^(2)+e_(L)^(2))" "therefore e_(R)=sqrt(e^(2)-e_(L)^(2))=132V` |
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| 130. |
What will be the phase difference between virtual voltage and virtual current, when the current in the circuit is wattlessA. `180^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
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Answer» Correct Answer - D Power `(P) = E_(rms) I_(rms) cos phi = 0` Since it is wattless current. But `E_(rms) !- 0, :. Cos phi = 0 or phi = 90^(@)`. |
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| 131. |
What is power dissipation in an a.c. circuit in which `E = 230 sin (omega t + pi//2) , I = 10 sin omega t`? |
| Answer» As phase diff. `= 90^(@)`, therefore average power dissipated `P = E_(v) I_(v) cos 90^(@) =` Zero | |
| 132. |
The primary and secondary voltage of an ideal step down transformer are 200 V and 25V respectively. The secondary is connected to a device, which draws a current of 2A. What is the current in the primary?A. 100 mAB. 150 mAC. 200 mAD. 250 mA |
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Answer» Correct Answer - D `(I_P)/(I_S) = (V_S)/(V_P) :. (I_P)/2 = 25/200` `:. I_(P) = 1/4 A = 250 mA`. |
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| 133. |
An alternating potential `E=E_(0) sin omega t` is applied to a series L-C circuit. What is the phase difference between the voltage across L and C?A. `(pi)/2`B. `pi`C. `(3 pi)/(2)`D. zero |
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Answer» Correct Answer - B For `L, E = E_(0) sin (omega t + pi//2)` For C, `E =E_(0) sin (omegat-pi//2)` `:.` phase difference = `pi`. |
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| 134. |
Show that time constant `(tau = RC)` of `R - C` circuit has the dimensions of time. |
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Answer» Time constant of RC circuit indicates how fast or how slow will be the charging and discharging of a capacitor through a resistor. A small value of `t = RC`, represents faster charging and discharging, whereas large value of `t = RC`, indicates slower charging and discharging of the capacitor. For a given circuit, time constant is fixed. We can optimise the value of time constant by choosing suitable value of R and C. |
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| 135. |
The figure shows an experimental plot discharging of a capacitor in an `RC` circuit. The time constant of this circuit lies between: A. 150 sec and 200 secB. 0 and 50 secC. 50 sec and 100 secD. 100 sec and 150 sec |
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Answer» Correct Answer - D From Fig. max p.d. `V_(0) = 25` volt Corresponding to `V = (V_(0))/(2) = (25)/(2) = 12.5 V`, `t = 100 s`. The discharging of a capacitor through resistance is governed by `V = V_(0) e^(-t//tau)` when `V = (V_(0))/(2) , (V_(0))/(2) = V_(0) e^(-t//tai)` `log_(e) 1 - log_(e) 2 = - (t)/(tau) log_(e) e = - (t)/(tau)` `tau = (t)/(log_(e) 2) = (100)/(0.693) = 144.3 s` Choice (d) is correct |
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| 136. |
As shown in figure, a metal rod completes the circuit. The circuit area is perpendicular to a magnetic field with `B = 0.15 T`. If the resistance of the total circuit is `3 Omega`, how large a force is needed to move the rod as indicated with a constant speed of 2 m/s? A. `3.75xx10^(-3)N`B. `2.75xx10^(-3)N`C. `6.57xx10^(-4)N`D. `4.36xx10^(-4)N` |
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Answer» Correct Answer - A Given, B = 0.15 T, l = `50xx10^(-2)` cm and v = 2 `ms^(-2)` `therefore` emf , e = Bvl `e=0.15xx2xx50xx10^(-2)=0.15` Current, `i=(e)/(R)=(0.15)/(3)=5xx10^(-2)` Force, F = Bil = `0.15xx5xx10^(-2)xx50xx10^(-2)` `=3.75xx10^(-3)N` |
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| 137. |
As shown in figure, a metal rod completes the circuit. The circuit area is perpendicular to a magnetic field with `B = 0.15 T`. If the resistance of the total circuit is `3 Omega`, how large a force is needed to move the rod as indicated with a constant speed of 2 m/s? A. `3.75xx10^(-3)N`B. `3.75xx10^(-2)N`C. `3.75xx10^(2)N`D. `3.75xx10^(-4)` |
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Answer» Correct Answer - A Induced current in the circuit `i=(Bvl)/(R )` Magneticforceactingonthewire `F_(m)=bil=B(Bvl)/(R ))l` `impliesF_(m)=(B^(2)vl^(2))/(R )` External force needed to move the rod with constant velocity `(F_(m))=(B^(2)vl^(2))/(R )=((0.15)^(2)xx(2)xx(0.5)^(2))/(3)` `=3.75xx10^(-3)N` |
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| 138. |
What are the SI units of magnetic field induction or magnetic flux density? |
| Answer» Correct Answer - Weber, Tesla. | |
| 139. |
A `0.5m` long metal rod `PQ` completes the circuit as shown in the figure. The area of the circuit is perpendicular to the magnetic field of flux density `0.15 T`. If the resistance of the total circuit is `3Omega` calculate the force needed to move the rod in the direction as indicated with a constant speed of `2 ms^(-1)` |
| Answer» `F=(B^(2)L^(2)v)/R=((0.15)^(2)(0.5)^(2)xx2)/3=0.00375 N` | |
| 140. |
In the figure given below a bar magnet moving towards the right or left includes an `emf` in the coii `(1)`and `(2)` .Find giving reason thedirections of the induced currents through the resistors `AB` and `CD` when the magnet is moving `(a)` towards the right and `(b)` towards the left |
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Answer» (a)When the magnet is moved towards right, the right end of coil `(1)` develps `S`-polarity and left end of coil `(2)` also develops south polarity i.e., current in both coils flow anticlockwise, in coil `(1)` current flows from `A` to `B` while in coil `(2)`,current flows from `D` to `C` (b)When the magnet is moved towards left, the right end of coil `(1)` develps `N`-polarity and left end of coil `(2)` also develops `N`- polarity i.e., current in both coils flow clockwise, in coil `(1)` current flows from `B` to `A` while in coil `(2)`,current flows from `C` to `D` |
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| 141. |
Define magnetic flux . Give its `SI` unit. `I` |
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Answer» Magnetic flux:Magnetic flux through any surface in magnetic field `vecB` is equal to the total number of magnetic lines of force crossing the surface `A`. `therefore phi=BA cos theta`.Its `SI` unit is weber.It is scalar quantity. |
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| 142. |
Define self -inductance and its `S.I.` unit. Derive an expression for self- industance of a long, air-cored solenoid of length , radius `r`, and having `N` number of turns. |
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Answer» The self inductance of a coil is equal to the induced `emf` set up in the coil, when the current passing through it change at the unit rate The `SI` unit of self inductance is henry. Self inductance of a long solenoid: Consider a long solenoid of length `l` number of turns `N` and radius.Suppose current `I` flows though it.Magnetic field set up in the coil is `B=(mu_(0)NI)/l` Flux through each turn=`BA=(mu_(0)NI)/l` But `phi=Li` `phi=(Nxxmu_(0)NI)/l =(mu_(0)N^(2)IA)/l` `therefore` Self inductance `L=phi/I=(mu_(0)N^(2)A)/l` |
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| 143. |
A boy peddles a stationary bicycle the pedals of the bicycle are attached to a 200 turn coil of area `0.10m^(2)`. The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.02 T perpendicular to the axis of rotation of the coil. The maximum voltage generated in the coil isA. `1.26V`B. `2.16V`C. `3.24V`D. `4.12V` |
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Answer» Correct Answer - A Here, `v=0.5 Hz. N=200,A=0.1m^(2)` and `B=0.02T` Maximum voltage generated is `epsi_(0)NBA(2piv)=200xx0.02xx0.1xx(2pixx0.5)=1.25V` |
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| 144. |
Kamla peddles a stationary bicyle, the pedals of the bicycle are attached to a 100 turns coil of area `0.10m^(2)`. The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.01T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil. |
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Answer» Here f=0.5 Hz, N=100, `A=0.1m^(2) and B=0.01T` Employing Equation `epsi=NBA omega sin omegat` `epsi_(0)=NBA (2piv)` `=100xx0.01 xx0.1 xx 2 xx 3.14 xx 0.5` `=0.314V` The maximum voltage is 0.314V. |
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| 145. |
A long solenoid with 15 turns per cm has small loop of area `2.0 cm^(2)` placed inside, normal to the axis of the soleniod. If current carried by the solenoid changes steadily from 2 A to 4 A in 0.1 s, what is the induced voltage in the loop, while the current is changing ? |
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Answer» Number of turns on the solenoid = 15 turns/cm = 1500 turns/m Number of turns per unit length, n = 1500 turns The solenoid has a small loop of area, A = 2.0 `cm^(2)` = `2 × 10^(−4) m^(2)` Current carried by the solenoid changes from 2 A to 4 A. `therefore`Change in current in the solenoid, di = 4 − 2 = 2 A Change in time, dt = 0.1 s Induced emf in the solenoid is given by Faraday’s law as: `e=(dphi)/(dt)`....(i) Where, `phi`= Induced flux through the small loop = BA ... (ii) B = Magnetic field =`mu_(0)ni`.....(iii) `mu_(0)` = Permeability of free space = `4pi×10^(−7)` H/m Hence, equation (i) reduces to: `e=d/(dt)(BA)` `=Amu_(0)nxx((di)/(dt))` `=2xx10^(-4)xx4pixx10^(-7)xx1500xx2/0.1` `=7.54xx10^(-6)` V Hence, the induced voltage in the loop is `=7.54xx10^(-6)` |
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| 146. |
Kamla peddles a stationary bicycle, the pedals of which are attached to a 100 turn coil of area `0.10 m^(2)`.The coil rotates at half a revolution in one second and it is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil ? |
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Answer» Here ν = 0.5 Hz, N =100, A = 0.1 `m^(2)` and B = 0.01 T. Employing Eq. (6.21) `epsi_(0)=NBA(2piv)` `=100xx0.01xx0.1xx2xx3.14xx0.5` `=0.314 V` The maximum voltage is 0.314 V. We urge you to explore such alternative possibilities for power generation. |
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| 147. |
The phase diffenernce between the alternating current and emf is `(pi)/(2)`. Which of the following cannot be the constiuent of the circuit?A. L aloneB. LCC. R,LD. C alone |
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Answer» Correct Answer - C L,C or LC can produce a phase difference of `(pi)/(2)` between alternating current and e.m.f. but it R is present, then the phases difference will lie between 0 and `(pi)/(2)` but it will not be `(pi)/2`. |
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| 148. |
In previous question calculate the average voltage per cycle of the alternating EMF. |
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Answer» Correct Answer - `(V_(0))/(sqrt(2))` |
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| 149. |
In an `AC` circuit, the mass value of the current `I_("rms")` is related to the peak current `I_(0)` asA. `I_(rms)=I_(0)//pi`B. `I_(rms)=I_(0)//sqrt(2)`C. `I_(rms)=piI_(0)`D. `I_(rms)=sqrt(2)I_(0)` |
| Answer» Correct Answer - B | |
| 150. |
AC measuring instruments measuresA. peak valueB. rms valueC. rms value of the currentD. rms value of the voltage |
| Answer» Correct Answer - B | |