1564 + Interview Questions in ELECTROMAGNETIC-INDUCTION IN GENERAL AWARENESS Page 27 InterviewSolution

1301.	An ac source of angular frequency `omega` is fed across a resistor R and a capacitor C in series. The current registered is I. If now the frequency of source is changed to `omega//3` (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of the reactance to resistance at the original frequency `omega`.A. `sqrt((3)/(5))`B. `sqrt((2)/(5))`C. `sqrt((1)/(5))`D. `sqrt((4)/(5))`
Answer» Correct Answer - A In `RC` circuit, `I_(upsilon) = (E_(upsilon))/(sqrt(R^(2) + ((1)/(omega C))^(2))` and `(I_(upsilon))/(2) = (E_(upsilon))/(sqrt(R^(2) + ((1)/((omega C)/(3)))^(2)` `:. 2 sqrt(R^(2) + ((1)/(omega C))^(2)) = sqrt(R^(2) + (9)/(omega^(2) C^(2))` or `3 R^(2) = (5)/(omega^(2) C^(2) = 5 X_(C )^(2)` `:. (X_(C ))/(R ) = sqrt((3)/(5))`

Discussion

1302.	A capacitor and a resistor are connected in series with an a.c. source. If the potential difference across C, R are 120 V and 90 V repectively, and if rms value of current is 3 A, calculate impedance and power factor of the circuit.
Answer» Here, `V_(L) = 120 V, V_(R ) = 90 V`, `I_(v) = 3 A, Z = ?, phi = ?` As voltages across R and L are `90^(@)` out of phase, `:.` resultant voltage, `V = sqrt(V_(L)^(2) + V_(R )^(2)) = sqrt(120^(2) + 90^(2)) = 150 V` `Z = (V)/(I_(v)) = (150)/(3) = 50 Omega` `tan phi = (X_(L))/(R) = (V_(L))/(V_(R )) = (120)/(90) = 1.33` `phi = tan^(-1) (1.33) = 53.1^(@)` In RL circuit, current lags behind the voltage.

Discussion

1303.	if the effective value of A.C in a circuit is 10A, then the peak value of current aA. `10 A`B. `1/(sqrt2) A`C. `14.14 A`D. `5A`
Answer» Correct Answer - C `I_(0) = sqrt(21_(rms)) = 1.414 xx 10 =14.14 A`.

Discussion

1304.	The equation of alternating of alternating current for a cuicuit is given by `I = 50 cos 100 pi t`. Calculate (i) frequency of a.c. applied,(ii) virtual value of current, (iii) value of current `1//600` s after it was maximum
Answer» Here, `I = 50 cos 100 pi t` Compare it with the standard form `I = I_(0) cos omega t`, we get `I_(0) = 50 A` `omega = 2 pi, v = (100 pi)/(2 pi) = 50 c//s` Virtual value of current, `I_(v) = (1)/(sqrt2) = (50)/(sqrt2)` At `t = 0, I = 50 cos 0 = 50 A = max`. At `t = (1)/(600) s` `I = 50 cos 100 pi xx (1)/(600) = 50 "cos" (pi)/(6)` `= 50 cos 30^(@) = (50sqrt3)/(2) = 25sqrt3 A`

Discussion

1305.	If the effective current in a 50 cycles a.c. cuicuit is 5 A, what is the peak value of current ? What is the current `1//600` sec. after it was zero ?
Answer» Here, `I_(v) = 5 A, I_(0) = sqrt2 I_(v) = 1.414 xx 5 = 7.07 A` From `I = I_(0) sin omega t` `I = 7.07 sin 2 pi xx 50 xx (1)/(600)` `= 7.07 (sin) (pi)/(6) = 7.07 xx (1)/(2) = 3.535 A`

Discussion

1306.	A capactior of `1 mu F` is connected ot an a.c. source of e.m.f `E = 250 sin 100 pi t`. Write an equation for instantaneous current through the circuit and given reading of a.c. ammeter connected in the circuit.
Answer» Here, `C = 1 mu F = 10^(-6) F` As `E = 250 sin 100 pi t` `:. E_(0) = 250 V, omega = 100 pi` The instantaneous current through the circuit is `I = I_(0) sin (omega t + pi//2) = (E_(0))/(1 // omega C) sin (omega t + pi // 2)` ` = 250 xx 100 pi xx 10^(-6) sin (100 pi t + pi // 2)` `I = 0.0786 sin (100 pi t + pi // 2)` Reading of a.c. ammeter `= I_(v)` `= (I_(0))/(sqrt 2) = (0.0786)/(sqrt 2) = 0.06 A`

Discussion

1307.	In a circuit, the instantaneous values of alternating current and voltages in a circuit is given by `I = (12)/(sqrt2) sin (100 pi t) A` and `E = (1)/(sqrt2) sin (100 pi t + (pi)/(3)) V`. The average power in watts consumed in the circui isA. `(1)/(4)`B. `(sqrt3)/(4)`C. `(1)/(2)`D. `(1)/(8)`
Answer» Correct Answer - D Here, on comparing with `I = I_(0) sin omega t ` and `E = E_(0) sin (omega t + phi)` we get `I_(0) = 1//sqrt2 A` `E_(0) = 1//sqrt2 V` `phi = (pi)/(3)` `E_(upsilon) = (E_(0))/(sqrt2) = (1 //sqrt2)/(sqrt2) = (1)/(2) V`, `I_(upsilon) = (I_(0))/(sqrt2) = (1 //sqrt2)/(sqrt2) = (1)/(2) A`, `cos phi = "cos" (pi)/(3) = (1)/(2)` `P_(av) = E_(upsilon) I_(upsilon) cos phi = ((1)/(2)) ((1)/(2)) ((1)/(2)) = (1)/(8) W`

Discussion

1308.	A coil is wound as a transformer of rectangular cross section. If all the linear dimension of the transformer are increased by a factor `2` and the number of turns per unit length of the coil remain the same, the self-inductance increased by a factor ofA. `16`B. `12`C. `8`D. `4`
Answer» Correct Answer - C Self-inductance `L=mu_(0)N^(2)A//l=mu_(0)n^(2)lA` where `n` is the number of turns per unit length and `N` is the total number of turns and `N=nl` In the given question `n` is same. `A` is increased `4` times and `l` is increased `2` times and hence `L` will be increased `8` times.

Discussion

1309.	`5 cm` long solenoid having `10` ohm resistance and `5mH` induced is joined to a `10` volt battery. At steady state the current through the solenoid in apmpere will beA. `5`B. `1`C. `2`D. zero
Answer» Correct Answer - B In steady state current passing through solenoid `i=(E)/(R )=(10)/(10)=1A`

Discussion

1310.	For a RLC series circuit , phasors of current i and applied voltage V`=V_(0)sin omegat` are shown in diagram at t=0, which of the following is/are CORRECT?A. At `=(pi)/(2omega)`, instantaneous power supplied by source is negative.B. From `0lttlt(2pi)/(3omega)`, average power supplied by source is positive.C. At `t=(5pi)/(6omega)`, instantaneous power supplied by source is negative.D. If `omega` is increased slightly, angle between the two phasors decreases.
Answer» Correct Answer - A::C

Discussion

1311.	A small signal voltage `V(t)=V_(0)sin omegat` is applied across an ideal capacitor `C`:A. over a full cycle, the capacitor does not consume any energy form the voltage sourceB. current `I (t)` is in phase with the voltage `V (t)`C. Current `(I (t)` leads the voltage `V (t)` by `180^(@)`D. current `I (t)` lags behind the voltage `V (t)` by `90^(@)`
Answer» Correct Answer - A Voltage applied acorss an ideal capcaitor `V = V_(0) sin omega t` Through capacitor, current leads the applied e.m.f. by a phase angle of `90^(@)` Average power over a complete cycle `P = VI cos phi = VI cos 90^(@) =` Zero. Choice (a) is correct.

Discussion

1312.	An oscillator circuit consists of an inductance of `0.5mH` and a capacitor of `20muF`. The resonant frequency of the circuit is nearlyA. `15.92Hz`B. `159.2Hz`C. `1592Hz`D. `15910Hz`
Answer» `v_(0)=(1)/(2pisqrt(LC))=(1)/(2xx3.14sqrt(5xx10^(-4)xx20xx10^(-6))` `v_(0)=(10^(4))/(6.28)=1592 Hz`

Discussion

1313.	A large circular coil of radius R and a small circular coil of radius r are put in vicinity of each other. If coefficient of mutual inductance of this pair equals 1 mH, what would be the flux linked with larger coil when a current of 0.5 A flows through the smaller coil? When would current in the smaller coil falls to zero, what would be its effect in the larger coil ?
Answer» Here, `M = 1 mH = 10^(-3) H, phi = ?` `I = 0.5 A, phi = MI = 10^(-3) xx 0.5 = 5 xx 10^(-4) Wb` When current in smaller coil falls to zero, an e.m.f. is induced in larger coil due to mutual induction.

Discussion

1314.	The current i in an inductionn coil varies with time `t` according to the graph shown in the figure. Which of the following grahs shows athe induced emf `(epsilon)` in the coil with time? B. C. D.
Answer» Correct Answer - C `e=L\|(di)/(dt)\|=L` (Slope of `i - t`graph) Initilly slope `=0implies e=0` Then in remaining two regions slopes are constant but of opposite signs. Hence, induced emfs are constants but of opposite signs.

Discussion

1315.	A coil of inductance `300mh` and resistance `2Omega` is connected to a source of voltage `2V`. The current reaches half of its steady state value inA. `0.05 s`B. `0.1 s`C. `0.15 s`D. `0.3 s`
Answer» Correct Answer - B The current at any instant is given by `l=l_(0)(1-e^(-Rt//L))` `l_(0)/2=l_(0)(1-e^(-Rt//L))` `1/2=(1-e^(-Rt//L))` `e^(-Rt//L)=1/2` `(Rt)/L=n2` `therefore t=L/R "ln" 2` `=(300 xx 10^(-3))/2xx0.693 =150xx0.693xx10^(-3)` `=0.10395 sec =0.1 sec`

Discussion

1316.	The current i in an inductionn coil varies with time `t` according to the graph shown in the figure. Which of the following grahs shows athe induced emf `(epsilon)` in the coil with time? A. B. C. D.
Answer» Correct Answer - C In Fig. when `i` is constant,` phi` constant Therefore, `e = 0` When `i` decrease, `phi` decreases, e.m.f. induced is positive constant. When `i` increases, `phi` increases, e.m.f. induced is negative constant. Choice (c ) is correct.

Discussion

1317.	An inductor `(L = 100 mH)`, a resistor `(R = 100 (Omega))` and a battery `(E = 100 V)` are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the point A and B. The current in the circuit 1 ms after the short circuit is A. 1AB. 1/eAC. eAD. 0.1A
Answer» Correct Answer - B

Discussion

1318.	An inductor `(L = 100 mH)`, a resistor `(R = 100 (Omega))` and a battery `(E = 100 V)` are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the point A and B. The current in the circuit 1 ms after the short circuit is A. `1 A`B. `1//e A`C. `e A`D. `0.1 A`
Answer» Correct Answer - B `l=l_(0)e^(-Rt//L)=1/eA`.

Discussion

1319.	An inductor (L = 100 mH), a resistor `(R = 100Omega)` and a battery (E = 100 V) are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit 1 ms after the short circuit is A. eAB. 0.1AC. 1AD. 1/eA
Answer» Correct Answer - D During decay of current, `i=i_(0)e^(-(Rt)/(L))=(E)/(R)e^(-(Rt)/(L))=(100)/(100)e^((100xx10^(-3))/(100xx10^(-3)))=(1)/(e)A`

Discussion

1320.	The reactance of an inductor at 50 Hz is `100Omega`. If the frequency is increased to 60 Hz, the reactance of the same inductor becomesA. `12Omega`B. `21 Omega`C. `1.2Omega`D. `120Omega`
Answer» Correct Answer - D `(X_(L1))/(X_(L2))=(f_(1))/(f_(2))=(50)/(60)` `therefore X_(L2)=(f_(2))/(f_(1))X_(L1)=120Omega`

Discussion

1321.	A square wire loop with sides 0.5 m is placed with its plane perpendicular to the magnetic field. The resistance of the loop is `5Omega`. The rate at which the magnetic induction should be changed so that a current of 0.1 A is induced in the loop isA. 4 T/sB. 3 T/sC. 2 T/sD. 1 T/s
Answer» Correct Answer - C `e=(dphi)/(dt)` `IR=(AdB)/(dt)` `(dB)/(dt)=(IR)/(A)=(0.1xx5)/(0.25)=2T//s`

Discussion

1322.	When a wheel with metal spokes 1.2 m long rotates in a magnetic field of flux density ` 5 xx 10^(-5)` T normal to the plane of the wheel, an e.m.f. of `10^(-2)` V is induced between the rim and the axle. Find the rate of rotation of the wheel.A. 44 rpsB. 88 rpsC. 22 rpsD. 11 rps
Answer» Correct Answer - A `e=Bpir^(2)f` `therefore f=(e)/(Bpir^(2))=(10^(-2))/(5xx10^(-5)xx3.14xx1.44)` `f=44`rps.

Discussion

1323.	The current in a `LR` circuit builds up to `(3)/(40th` of its steady state value ion `4s`. The time constant. Of this circuit isA. `(1)/("in"2)s`B. `(2)/("in"2)s`C. `(3)/("in"2)s`D. `(4)/(I"n" 2)s`
Answer» Correct Answer - B We know that `i=i_(0)[1-e^((-Rt)/(L))]or (3)/(4)i_(0)=i_(0)[1-e^(-t//tau]` (where `tau=(L)/(R )=`time constant) `(3)/(4)=1-e^(-t//tau)ore^(-t//tau)=1-(3)/(4)=(1)/(4)` `e^(t//tau)=4or(t)/(tau)=In4` `impliestau=(t)/(In 4)=(4)/(2In2)impliestau=(2)/(In2)sec`.

Discussion

1324.	A magnetic flux of `8xx10^(-4)` weber is linked with each turn of a 200 - turn coil when thereis an electric current of 4 A in it. Calculate the self inductance of the coil.
Answer» `"Total flux linked "phi=200xx8xx10^(-4)=0.16" weber"` `"Now inductance L"=(phi)/(i)=(0.16)/(4)=0.04-40mH`

Discussion

1325.	The self inductance of a coil is L . Keeping the length and area same, the number of turns in the coil is increased to four times. The self inductance of the coil will now beA. `(1)/(4)L`B. LC. 4 LD. 16 L
Answer» Correct Answer - D Self-inductance of coil is directly proportional to squre of number of turns in the coil, i.e., `L propN^(2)` `therefore" "(L_(1))/(L_(2))=(N_(1)^(2))/(N_(2)^(2))` `therefore" "(L)/(L_(2))=(N^(2))/((4N)^(2))` `therefore" "L_(2)=16L`

Discussion

1326.	A coil of area `80` square cm and 50 turns is rotating with `2000` revolution per minut about an axis perpendicular to a magnetic field field of `0.05` Telsa. The maximum value of the e.m.f. developed in it isA. `(2pi)/(3)V`B. `(4pi)/(3)V`C. `(5pi)/(3)`D. `(pi)/(3)V`
Answer» Correct Answer - B `e_(0)=nABomega=(4pi)/(3)V.`

Discussion

1327.	The self inductance of a coil is L . Keeping the length and area same, the number of turns in the coil is increased to four times. The self inductance of the coil will now beA. 4 LB. 8 LC. 16 LD. 12 L
Answer» Correct Answer - C `L propn^(2) implies (L_(2))/(L_(1))=(n_(2)^(2))/(n_(1)^(2))" "therefore L_(2)=16L`

Discussion

1328.	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
Answer» `B=(mu_(0)iR^(2))/(2(R^(2)+X^(2))^(3//2))` `B=(mu_(0)iR^(2))/(2(R^(2)+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=NBA cos45^(@)` `=2(mu_(0)i)/(16R)a^(2)1/sqrt2` `phi=(mu_(0)ia^(2))/(8sqrt2R)` `M=phi/i` `M=(mu_(0)a^(2))/(2^(7//2)R)=(mu_(0)a^(2))/(2^(P//2)R)` `P=7`

Discussion

1329.	In the circuit shown in fig. potential diff. across L, C and and R are given. Find the e.m.f. of the source and calculate power factor of the circuit.
Answer» Here, `V_(L) = 100 V, V_(R ) = 80 V`, `V_(C ) = 40 V, V = ?` `V = sqrt(V_(R )^(2) + (V_(L) - V_(C ))^(2)) = sqrt(80^(2) + (100 - 40)^(2))` `=100 V` Power factor, `cos phi = (V_(R ))/(V) = (80)/(100) = 0.8`

Discussion

1330.	If the current flowing through a coil is reduced by 50% then the energy in the coilA. will not increasedB. will not changeC. will be decreased by 20%D. will be decreased by 75%
Answer» Correct Answer - D `(E_2)/(E_1)= (1/2L((I)/(2))^(2))/(1/2LI^(2)) =1/4` `E_(2) = 1/4 E_(1) :. E_(1)-E_(2) = 3/4 E_(1)`.

Discussion

1331.	An electromagnet has stroed 648 J of magnetic energy, when a current of 9 A exists in its coils. What average e.m.f. is induced if the current is reduced to zero in 0.45
Answer» Her, `E = 648 J, I = 9 A, e =?` `dI = 9 - 0 = 9 A, dt = 0.45 s` From `E = (1)/(2) LI^(2)` `648 = (1)/(2) xx L (9)^(2)`, `L = (648 xx 2)/(9 xx 9) = 16 H` As `e = (L dI)/(dt) :. e = (16(9))/(0.45) = 320 V`

Discussion

1332.	A conducting loop is placed in a uniform magnetic field with its plane perpendicular to the field with its plane perpendicular to the field. An emf is induced in the loop if (i) it is translated (ii) it is rotated about its axis (iii) it is rotated about a diameter (iv) it is deformedA. It is translatedB. It is rotated about its axisC. Is rotated about a diameterD. It is deformed
Answer» Correct Answer - C::D emf will be induced when loop is rotated about a diameter and if deformed because that will changed the magnetic flux and emf will be induced.

Discussion

1333.	A force of 10N is required to move a conducting loop through a non-uniform magnetic field to 2m/s. The rate of production of internal energy in loop is:A. 25WB. 5WC. 10WD. 20W
Answer» Correct Answer - C

Discussion

1334.	Two plane circular coils P and Q have radii `r_(1)` and `r_(2)`, respectively, `(r_(1) lt lt r_(2))` and are coaxial as shown in fig. The number of turns in P and Q are respectively `N_(1)` and `N_(2)`. If current in coil Q is varied steadily at a rate x ampere/second then the induced emf in the coil P will be approximatery - A. `mu_(0)N_(1)N_(2) pi r_(1)^(2)`B. `mu_(0)N_(1)N_(2)pi r_(1)^(2) x`C. `mu_(0)N_(1)N_(2) pi r_(1)^(2) x// 2r_(2)`D. zero
Answer» Correct Answer - C `phi=Mi` `e=(Mdi)/(dt)=Mx=(mu_(0)N_(1)N_(2) pi r_(1)^(2)x)/(2r_(2))`

Discussion

1335.	The magnetic flux of `phi` (in weber) in a closed circuit of resistance `3 ohm` varies with time `t ` (in second according to the equation `phi = 2 t^(2) - 10 t` at ` t = 0.25 s`?
Answer» Correct Answer - 3 Here, `R = 3 Omega, phi = 2 t^(2) - 10 t +3, I = ?, t = 0.25 s` `e = - (d phi)/(dt) = - (d)/(dt) (2 t^(2) - 10 t + 3)` `e = - 4 t + 10` At `t = 0.25 s, e = - 4 xx 0.25 + 10 = 9` volt `I = (e)/(R ) = (9)/(3) = 3 A`

Discussion

1336.	Does it imply that power dissipated in an a.c. circuit is zero at resonance ?
Answer» No , power dissipated at resonance is maximum.

Discussion

1337.	In the given circuit, let `i_(10)` be the current drawn battery at time `t=0` and `i_(20)` be steady current at `t=oo` then the ratio `(i_(1))/(i_(2))` is A. `1.0`B. `0.8`C. `1.2`D. `1.5`
Answer» Correct Answer - B At `t=0`, inductor behave as open circuit so `i_(1)` `=(10)/(10)=1 A` At `t=oo`, inductor behave as short circuit, so `i_(2)` `=(10)(8)=(5)/(4)A` Hence, `(i_(1))/(i_(2))=(1)/(5//4)=(4)/(5)=0.8`

Discussion

1338.	Two coils P and S kept very close to each other. When the current in P changes by 10A, the magnetic flux in S changes by 1.5 weber. The mutal inductance of the coil isA. 1.5 HB. 2.5 HC. 0.15 HD. 0.8 H
Answer» Correct Answer - C `phi_(S) = MI_(P) :. 1.5 = M xx 10 :. M = 0.15 H`.

Discussion

1339.	A 200 km long telegraph wire has capacity of `0.14 mu F//km`. If it carries an alternating current of `50 kc//s`, what should be the value of inductance required to be connected in series so that impedance is minimum ?
Answer» Here, `l = 200 km` Capacity ` = 0.014 mu F//km` `:.` Total capacity of telegraph wire `C = 0.014 xx 200 mu f = 2.8 xx 10^(-6) F` `v = 50 kC//s = 50 xx 10^(3) C//s, L = ?` Impedance will be minimum only at resonance, when ` v = (1)/(2 pi sqrt(LC ))` `:. L = (1)/(4 pi^(2) v^(2) C)` `= (1)/(4 xx (22)/(7) xx (22)/(7) xx (50 xx 10^(3))^(2) xx 2.8 xx 10^(-6))` `L = 0.36 xx 10^(-5) H`

Discussion

1340.	If the P.D. across the inductor `(3mH)` is the same as that across the condenser `(30 mu F)` in a series R-L-C circuit, then the frequency of the applied e.m.f. isA. 180 HzB. 500 HzC. 890 HzD. 5 kHz
Answer» Correct Answer - B `f=(1)/(2pisqrt(LC))` `=(1)/(2xx3.14xxsqrt(3xx10^(-3)xx30xx10^(-6)))` `=500Hz`

Discussion

1341.	If the P.D. across the inductor `(3mH)` is the same as that across the condenser `(30 mu F)` in a series R-L-C circuit, then the frequency of the applied e.m.f. isA. 180 HzB. 530 HzC. 890 HzD. 5 KHz
Answer» Correct Answer - B When the P.D. across L=P.D. across C in a series L,C,R circuit, we get resonance. At resnance, `X_(L)=X_(C)`. `:. omega L = 1/(omega C)` and the resonant frequency `f = 1/(2 pisqrt(LC))` `=1/(2 xx 3.14) xx (1)/(sqrt(3 xx 10^(-3)xx30xx10^(-6))) = 530 Hz`.

Discussion

1342.	Statement-1 : An inductor cannot have zero resistance. Statement-2 : This is because inductor has to be made up of some material, which must have some resistance.A. AB. BC. CD. D
Answer» Correct Answer - A Both the statements are ture, and statement-2 is correct explanaiton of statement-1.

Discussion

1343.	A straight wire of length `L` is bent into a semicircle. It is moved in a uniform magnetic field with speed `v` with diameter perpendicular to the field. The induced emf between the ends of the wire is A. BLvB. 2BLvC. `2piBLv`D. `(2BvL)/(pi)`
Answer» Correct Answer - d

Discussion

1344.	A rod of mass m length l can rotate without friction about the centre of a vertical ring . There is a unifrom mafgnetic filed B into the plane of the ring . A variable emf `epsilon` is applied between the centre and the rotating end of the rod . Caulcate the current which keeps the rod rotating with unifrom speed `omega` and the emf requried to maintain the required current
Answer» Correct Answer - `i=(mg cos omega t)/(Bl), epsi=(1)/(2) Bomegal^(2)+(mg R cos omegat)/(Bl)`

Discussion

1345.	A wire with a resistance p per unit length is bent in the from of the letter A of vertical angle `2alpha` . There is a magnetic field B perpendicular into the plane of the letter . Calculate the current flowing in the loop when the cross-piece cut moves down at a constant speed v . Assume that it maintains contact with the sides as it moves down .
Answer» Correct Answer - `i=(Bv)/((1+ cos alpha)rho)`

Discussion

1346.	A transformer is used to step down a.c.voltage. What appliance do you use to step down d.c. voltage ?
Answer» For stepping down d.c. voltage, we may use an ohmic resistance only such as in potential dividing arrangement.

Discussion

1347.	Name any appliance that can step down d.c. voltage.
Answer» A pure resistor can be used to step down d.c. voltages.

Discussion

1348.	Current is switched on in a circuit contaning a resistance of `10 Omega` and an inductance of 0.8 H. Calculate the time taken by the current to grow to half its maximum value.
Answer» Here, `R = 10 Omega, L = 0.8 H`, `t = ?, I = I_(0) //2` As `I = I_(0) (1 - e^(-R//Lt)) = I_(0) //2` `:. (1)/(2) = 1 - e^(-Rt//L)` or `e^(-Rt//L) = 1 - (1)/(2) = (1)/(2)` `(Rt)/(L) log_(e) e = log_(e) 1 - log_(e) 2 = 0 - 0.693` `t = + 0.693 (L)/(R ) = (0.693 xx 0.8)/(10) = 0.055 s`

Discussion

1349.	Predict the direction of induced current in the situation described by follow (f)
Answer» The magnetic field lines due to the current carrying wire.are in the plane of the loop. Hence, no induced current is produced in the loop (because no flux lines crosses the area of loop).

Discussion

1350.	The circuit shown in Fig. is switched on at t = 0. Calculate the time at which current in `R_(2)` becomes half the steady value of current. Also, calculate the energy stored in the inductor at that time.
Answer» As is known, final steady current in the circuit does not depend on L. `:. I_(0) = (E)/(R_(1) + R_(2)) = (3)/(2 + 3) = 0.6 A` The growth of current in RL circuit is given by Helmholtz equation `I = I_(0) [1 - e^((-R//L) t)]` For `I = I_(0)//2`, we have to find t `:. (I_(0))/(2) = I_(0) [1 - e^((-R//L)t)]` `(1)/(2) = 1 - e^(-Rt//L)` `e^(-Rt//L) = 1 - (1)/(2) = (1)/(2)` `- (Rt)/(L) = log_(e) 1 - log_(e) 2 = 0 - 2.3026 xx 0.3010` `= - 0.6931` `t = 0.931 xx (L)/(R ) = (0.931 xx 10 xx 10^(-3))/(2 + 3)` `t = 1.386 xx 10^(-3) s` Now, `I = I_(0)//2 = 0.6//2 = 0.3 A` `:.` Energy store in the inductor `U = (1)/(2) LI^(2) = (12)/(2) (10 xx 10^(-3)) (0.3)^(2)`

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