1564 + Interview Questions in ELECTROMAGNETIC-INDUCTION IN GENERAL AWARENESS Page 6 InterviewSolution

251.	A conducting wire of 100 turns is wound over and near the centre of a solenoid of 100 cm length and 2 cm radius having 600 turns. Calculate mutual inductance of two coils.
Answer» Here, `N_(2) = 100, l = 100 cm = 1 m`, `r = 2 cm = 2 xx 10^(2) m, N_(1) = 600 , M = ?` `M = (mu_(0) N_(1) N_(2) A)/(l) = (mu_(0) N_(1) N_(2) pi r^(2))/(l)` `= (4 pi xx 10^(-7) xx 600 xx 100 xx 3.14 (2 xx 10^(2))^(2))/(1)` `M = 9.47 xx 10^(-5) H`

Discussion

252.	A 2 m long solenoid with diameter 4 cm and 2000 turns has a secondary of 1000 turns wound closely near its mid point. Calculate the mutual inductance between the two coils.
Answer» Here, `l = 2 m, r = (4)/(2) = 2 cm 2 xx 10^(2) m` `N_(1) = 2000, N_(2) = 1000, M = ?` `A = pi r^(2) = pi (2 xx 10^(-2)) m^(2) = 4 pi xx 10^(-4) m^(2)` `M = (mu_(0) N_(1) A)/(l) = (4 pi xx 10(-7) xx 2000 xx 1000 xx 4 pi xx 10^(-4))/(2) = 1.5 xx 10^(-3) H`

Discussion

253.	A solenoid of 2000 turns is wound over a length of 0.3 m. The area aof cross section is `1.3 xx 10^(-3) m^(2)`. Around its central seciton, a coil of 300 turns is closely wound. If an initial current of 2 A is reversed in 0.25 s, find the e.m.f induced in the coil .
Answer» Here, `N_(1) = 2000m l = 0.3 m`, `A = 1.2 xx 10^(-3) m^(2), N_(2) = 300` `dI = 2 - (-2) = 4 A, dt = 0.25 s, e = ?` `M = (mu_(0) N_(1) N_(2) A)/(1)` ` (4 pi xx 10^(-7) xx 2000 xx 300 xx 1.2 xx 10^(-3))/(0.3)` `=3 xx 10^(-3) H` `\|e\| = M (dI)/(dt) = 3 xx 10^(-3) xx (4)/(0.25) = 0.048 V`

Discussion

254.	The opposition offered by ohmic and non ohmic components isA. inductive reactanceB. capacitive reactanceC. impedanceD. all of these
Answer» Correct Answer - C

Discussion

255.	A carspark coil developes an induced e.m.f. of 40000 V in the secondary when when current in primary changes form 4 A to zero in `10 mu s`. What is the mutual inductance of the coil ?
Answer» Here, `e = 40000 V, dI = 0 - 4 = - 4 A`. `dt = 10 mu s = 10^(-5) s, M = ?` As `\|e\| = M (dI)/(dt)`, `M = \|e\| (dt)/(dt) = (40000 xx 10^(-5))/(4) = 0.1 H`

Discussion

256.	Over a solenoid of 50 cm length and 2 cm radius having 500 turns. Is wound another wire of 50 turns near the centre. Calculate mutual inductance of the two coils. If currents in primary changes From 0 to 5 a in 0.02 s, what is the emf induced in secondary coil ?
Answer» Here, `l = 50 cm = (1)/(2) m, r = 2 cm = 2 xx 10^(-2) m`, `N_(1) = 500 , N_(2) = 50, M = ?, dI = 5 - 0 = 5 A` `dt = 0.02 s, e= ?` `M = (mo_(0) N_(1) N_(2) A)/(l)` `= (4 pi xx 10^(-7) xx 500 xx 50 xx 3.14 (2 xx 10^(-2))^(2))/(1//2)` `= 7.896 xx 10^(-5) H` `e = (M dI)/(dt) = 7.896 xx 10^(-5) xx (5)/(0.02)` `= 19.74 xx 10^(-3) V = 19.74 mV`

Discussion

257.	Calculate current drawn by primary coil of a transformer, Which steps down 200 V to 20 V to operate a device of 20 ohm resistance. Assume efficiency of transformer 80 %.
Answer» Here, `I_(p) = ?, E_(p) = 200 V`, `E_(s) = 20 V, R_(s) = 20 ohm` `eta = 80% eta = (E_(s) I_(s))/(E_(p) I_(p)) = (E_(s) (E_(s)//R_(s)))/(E_(p) I_(p))` `(80)/(10) = (20 xx 20)/(20 xx 200 I_(p)) = (1)/(10 I_(p))` `I_(p) = (100)/(800) = 0.125 A`

Discussion

258.	The graphs shown in Fig. represent the variation of opposition offered by the circuit element of the flow of a.c. with frequency of applied e.m.f. Iddntify the circuit element .
Answer» In Fig, opposition to current does not vary with frequency. The circuit element must be a pure resistor. In Fig, opposition to current increses directly with applied frequency. The circuit element must be a pure inductor `(X_(L) = omega L = 2 pi v L , X_(L) prop v)`.

Discussion

259.	Show that the sum of instantaneous currents during growth and decay of current in LR circuit is independent of time.
Answer» During growth of current in LR circuit, `I_(g) = I_(0) (1 - e^(-Rt//L))` and during decay of current in LR circuit, `I_(d) = I_(0) e^(-Rt//L)` Sum of the currents at any instant t, `I_(g) + I_(d) = I_(0) (1 - e^(-Rt//L) + e^(-Rt//L)) = I_(0)` which is constant, and is independent to time.

Discussion

260.	A 66 % efficient transformer is working on 110 V and 2.2 kW power.If current in secondary coil is 6.0 A, calculate (i) current in primary coil and (ii) voltage across secondary coil.
Answer» Here, `eta = 66//100, E_(p) = 110 V`, `P_(p) = 2.2 kW = 2200 W` `I_(s) = 6.0 A, I_(p) = ?, E_(s) = ?` As `P_(p) = E_(p) I_(p) :. I_(p) = P_(p)//E_(p) = (2200)/(110) = 20 A` As `eta = (E_(s) I_(s))/(E_(p) I_(p)) :. (66)/(100) = (E_(s) xx 6)/(2200)` `E_(s) = (66 xx 2200)/(100 xx 6) = 242 V`

Discussion

261.	Fig. show a series LCR circuit with `L = 0.1 H l, X_(C ) = 14 Omega` and `R = 12 Omega` connected to a 50 Hz, 200 V source. Calculate (i) current in the circuit and (ii) phase angle between current and voltage. Take `pi = 3`
Answer» Here, `R = 12 Omega, X_(C ) = 14 Omega, L = 0.1 H` `E_(v) = 200 V, v = 50 Hz` `X_(L) = omega L = 2 pi v L = 2 xx 3 xx 50 xx 0.1 = 30 Omega` `Z = sqrt(R^(2) + (X_(L) - X_(C))^(2)) = sqrt(12^(2) + (30 - 14)^(2)) = 20 Omega` `I_(v) = (E_(v))/(Z) = (200)/(20) = 10 A` As `X_(L) gt X_(C )`, current lags behind the applied voltage by phase angle `phi`, where `tan phi = (X_(L) = X_(C ))/(R ) = (30 - 14)/(12) = 1.3333` `phi = tan^(-1) (1.3333) = 53.1^(@)`

Discussion

262.	A transformer has an efficiency of 80%. It works at 4 kW, 100 V. If the secondary voltage is 240 V, Calculate the primary and secondary currents.
Answer» Here, `eta = 80%, P_(i) = 4 kW = 4000 watt, E_(p) = 100 V, E_(s) = 240 V, I_(p) = ? I_(s) = ?` As `P_(i) = E_(p) I_(p) :. I_(p) = (P_(i))/(E_(p)) = (4000)/(100) = 40 A` As `eta = (E_(s) I_(s))/(E_(p) I_(p)) :. (80)/(100) = (240 I_(s))/(4000) :. I_(s) = (80 xx 4000)/(100 xx 240) = 13.3 A`

Discussion

263.	In an oscillating LC circuit the maximum charge on the capacitor is Q. The charges on the capacitor when the energy is stored equally between the electric and magnetic field isA. `(Q)/(2)`B. `(Q)/(sqrt(2))`C. `(Q)/(sqrt(3))`D. `(Q)/(3)`
Answer» Correct Answer - B `q=q_(0)cosomegat` , `omega=(1)/(sqrt(LC))` `=Qcosomegat` ltbr. `(q^(2))/(2C)=(1)/(2).(Q^(2))/(2C)` implies `q=(Q)/(sqrt(2))`

Discussion

264.	In an oscillating LC circuit the maximum charge on the capacitor is Q. The charges on the capacitor when the energy is stored equally between the electric and magnetic field isA. `Q/(sqrt(2))`B. QC. `Q/(sqrt(3))`D. `Q/2`
Answer» Correct Answer - A Let Q be the maximum charge on the capacitor and the maximum energy `E = 1/2 (Q^2)/C`. When it is equally shared between the eleclric and magnetic fields, the energy stored in the capacitor = `1/2 [1/2 (Q^2)/(C)]` In this case , If q is the charge on the capacitor, then `1/2 (q^2)/(C) = 1/2 [1/2 (Q^2)/C]`. `q^(2) = (Q^2)/C :. q=Q/(sqrt2)`.

Discussion

265.	What causes sparking in the switchs when light is put off ?
Answer» Large induced e.m.f. at break causes the sparking

Discussion

266.	A solenoid `S_1` is placed inside another solenoid `S_2` as shown in The radii of he inner and the outer solenoids are `r_1` and `r_2` respectively and the numbers of turns per unit length are `n_1` and `n_2` respectively. Consider a length I of each solenoid. Calculate the mutual inductance betwwen them.
Answer» Correct Answer - A (1) Magnetic field due to outer solenoid `B_(2)=mu_(0)n_(2)i` Flux through solenoid (1) `phi=B_(2).(n)_(1)l)(pir_(1)^(2))=mu_(0)n_(2)ilpir_(1)^(2)=Mi` `M=mu_(0)pin_(1)n_(2)r_(1)^(2)l`

Discussion

267.	Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area `A=10 cm^(2)` and length =20cm. If one of the solenoid has 300 turns and the other 400 turns, their mutual indcutance isA. `4.8pixx10^(-4)H`B. `4.8pixx10^(-5)H`C. `2.4pixx10^(-4)H`D. `4.8pixx10^(4)H`
Answer» Correct Answer - C `M=mu_(0)pir_(1)^(2)n_(1)n_(2)l` `=(mu_(0)AN_(1)N_(2))/(l)` `=(4pixx10^(-7)xx10^(-4)xx300xx400)/(0.2)` `=2.4pixx10^(-4)H`

Discussion

268.	Two coils of self-inductance `2mH` and `8 mH` are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coil isA. `10mH`B. `6mH`C. `4mH`D. `16mH`
Answer» Correct Answer - C When the total flux associated with one coil links with the other, i.e., a case of maximum flux linkage, then `M_(12)=(N_(2)phi_(B)_(2))/(i_(1))` and `M_(21)=(N_(1)phi_(B)_(1))/(i_(2))` similarly, `L_(1)N_(1)phi_(B)_(1))/(i_(1))` and `L _(2)=(N_(2)phiB_(2))/(i_(2))` if all the flux of coil `2` links coil `1` and vice versa, then `phi_(B)_(2)=phi_(B)^(1)` Since, `M_(12)=M_(21)=M`, hence we have `m_(12)M_(12)=M^(2)=(N_(1)N_(2)phi_(B)_(1)phi_(B)_(2))/(i_(1)i_(2)) =l_(1)L_(2)` `M_(max)=sqrt(L_(1)L_(2))` Given, `L_(1)=2mH,L_(2)=8mH` `M_(max)=sqrt(2xx8)=sqrt(16)=4mH`

Discussion

269.	Two coils of self-inductance `2mH` and `8 mH` are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coil isA. `4mH`B. `16mH`C. `10mH`D. `6mH`
Answer» Correct Answer - A `M=sqrt(L_(1)L_(2))=sqrt(2xx8)=4mH`

Discussion

270.	Two coils of self-inductance `2mH` and `8 mH` are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coil isA. 16 mHB. 10 mHC. 6 mHD. 4 mH
Answer» Correct Answer - D

Discussion

271.	Two coils of self-inductance `2mH` and `8 mH` are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coil isA. 4 mHB. 16 mHC. 10 mHD. 6 mH
Answer» Correct Answer - A Mutual inductance between coils is `M=Ksqrt(L_(1)L_(2))` `rArr" "M=1sqrt(2xx10^(-3)xx8xx10^(-3))" "[thereforeK=1]` `=4xx10^(-3)="4 mH"`

Discussion

272.	The self inductance of a closely wound coils of 100 turns is 5 mH . What is the flux throught the coil when the current in it is 10 mA ? [Hint : See exerices 5]
Answer» Correct Answer - `5 xx 10^(-5) wb`

Discussion

273.	The self inductance of a coil having 400 turns is 10 mH. The magnetic flux through the cross section of the coil corresponding to current 2 mA isA. `4xx10^(-5)Wb`B. `2xx10^(-3)Wb`C. `3xx10^(-5)Wb`D. `8xx10^(-3)Wb`
Answer» Correct Answer - A Here, `N=400,L=10 mH=10xx10^(-3)H` `I=2mA=2xx10^(-3)A` Total magnetic flux linked with the coil, `phi=NLI=400xx(10xx10^(-3))xx2xx10^(-3)=8xx10^(-3)Wb` Megnetic flux through the cross-section of the coil = Magnetic flux linked with each turn `(phi)/(N)=(8xx10^(-3))/(200)=4xx10^(-5)Wb`

Discussion

274.	The self inductane of coil of 400 turns is 8 mH. If current of 5 mA flows in it, then flux associated with the coil isA. `(mu_(0)//4pi)`B. `mu_(0)`C. `(mu_(0))/(100pi)`D. `(4pi//mu_(0))`
Answer» Correct Answer - C `phi=LI` `=8xx10^(-3)xx5xx10^(-3)` `=(mu_(0))/(100pi)`

Discussion

275.	A coil of self inductance 80 mH carries a current of 2A. What is the energy stored in the coil?A. 0.1 JB. 1.6 JC. 0.16 JD. 0.4 J
Answer» Correct Answer - C Energy = `1/2 LI^(2) = 1/2 xx 80 xx 10^(-3) xx 4 = 0.16 J`.

Discussion

276.	A coil of self inductance 20 mH, having 50 turns, carries a current of 300 mA. If the area of the coil is `2cm^(2)`, the magnetic induction at the centre of the coil isA. `7.5xx10^(-3)T`B. `7.5xx10^(-2)T`C. `6xx10^(-1)T`D. `0.5T`
Answer» Correct Answer - C `B=(mu_(0)nI)/(2a)` Now `A=pia^(2)" "therefore a=sqrt((A)/(pi))` `=(4xx10^(-7)xx50xx300xx10^(-3))/(2xxsqrt((2xx10^(-4))/(pi)))` `=0.6T`

Discussion

277.	A long solenoid has `1000` turns. When a current of `4A` flows through it, the magnetic flux linked with each turn of the solenoid is `4xx10^(-3)Wb`. The self-inductance of the solenoid isA. ` 3 H`B. ` 2 H`C. `1 H`D. `4 H`
Answer» Correct Answer - C Here, `N = 100m I = 4A, phi_(turn) = 4 xx 10^(-3) = 4 Wb`, `L = ?` Total flux, `phi = N phi_(turn) = 1000 xx 4 xx 10^(-3) = 4 Wb` From `phi = LI, L = (phi)/(I) = (4)/(4) = 1 H`

Discussion

278.	magnetic flux of 20 `mu` Wb is linked with a coil when current of 5 mA is flown through it. What is the self induction of the coil?
Answer» Here, `phi = 20 mu Wb = 20 xx 10^(-6) Wb` `I = 5 mA = 5 xx 10^(-3) A, L = ?` From `phi = LI, L = (phi)/(I) = (20 xx 10^(-6))/(5 xx 10^(-3))` `= 4 xx 10^(-3) H = 4 mH`

Discussion

279.	If a change in current of `0.01 A` in one coil produces change in magnetic flux of `1.2xx10^(-2)Wb` in the other coil, then the mutual inductance of the two coils in henriesA. `0`B. `0.5`C. `1.2`D. `3`
Answer» Correct Answer - C `varphi=MiimpliesM=(1.2xx10^(-2))/(0.01)=1.2H`

Discussion

280.	The coefficient of mutual inductance when magnetic flux change by `2xx10^(-2)Wb` and current changes by `0.01A`, will beA. `4H`B. `3H`C. `8H`D. `2H`
Answer» Correct Answer - C Coefficient of mutual induction of two coils is equal to the number of magnetic flux `(phi)` linkages in one coil when a unit current (i) flows in the other. therefore, `M=(phi)(i)` Given, `phi=2xx10^(-2)Wb,i=0.01A` `:. M=(2xx10^(-2))/(0.1)=2H`

Discussion

281.	Number of turns in a coil are 100. When a current of 5A is flowing through the coil, the magnetic flux is `10^(-6)Wb.` Find the self induction.
Answer» Self inductance, `L= (nphi)/(i)` number of turns, n=100, magnetic flux, `phi =10^(-6)" "Wb, Current, i=5A"` `L=(100 xx 10^(-6))/(5)=20xx10-6 =20muH therefore "Self inductance ,L"=20muH`

Discussion

282.	A coil of cross sectional area 100 `cm^(2)` is placed in the magnetic field, which changes to `4xx10^(-2)Wb//cm^(2)` within 5s. What will be the current across `5Omega` resistance?A. 0.016 AB. 0.16 AC. 1.6 AD. 16.0 A
Answer» Correct Answer - B `e=A(dB)/(dt)` `=(100xx10^(-4)xx4xx10^(2))/(5)=0.8V` `I=(e)/(R)=(0.8)/(5)=0.16A`

Discussion

283.	When current flowing in a coil changes from 3A to 2A in one millisecond, 5 volt emf is induced in it. The self-inductance of the coil will be-A. zeroB. 5khC. 5HD. 5 mH
Answer» Correct Answer - D `bot=-V/((Deltai)/(Deltat))=+5/(10^(3))=+5xx10^(-3) H =5mH`

Discussion

284.	Two coils have mutual inductance of 1.5 henry. If current in primary circuit is raised to 5 ampere in one millisecond after closing the circuit, what is the e.m.f. Induced in the secondary ?
Answer» Here, `M = 1.5`, henry, `dI = 5` apmere `dt = 1` milli sec. ` = 10^(-3) sec., e = ?` As `e = M (dI)/(dt) :.e = 1.5 xx (5)/(10^(-3)) = 7.5 xx 10^(-3) V`

Discussion

285.	A coil of area `0.1m^(2)` has 500 tums. After placing the coil in a magnetic field of strength `4xx10^(-4)Wb//m^(2)` it is rotated through `90^(@)` in 0.1 s. The average emf induced in the coil isA. 0.2 voltB. 0.1 voltC. 0.05 voltD. 0.012 volt
Answer» Correct Answer - A

Discussion

286.	In a magnetic field of `0.05T`, area of a coil changes from `101 cm^(2)` to `100 cm^(2)` without changing the resistance which is `2Omega`. The amount of charge that flow during this period isA. `2.5xx10^(-6)C`B. `2xx10^(-6)C`C. `10^(-6)C`D. `8xx10^(-6)C`
Answer» Correct Answer - A `therefore` Magnetic flux, `phi=B.A` `therefore` Change in flux, `dphi=B.dA` `=0.05(101-100)xx10^(-4)=5xx10^(-6)Wb` `therefore" Change, dq"=(dphi)/(R)=(5xx10^(-6))/(2)=2.5xx10^(-6)C`

Discussion

287.	In a uniform magneitc field of induced B a wire in the form of a semicircle of radius r rotates about the diameter of hte circle with an angular frequency `omega`. The axis of rotation is perpendicular to hte field. If the total resistance of hte circuit is R, the mean power generated per period of rotation isA. `(Bpir^(2)omega)/(2R)`B. `((Bpir^(2)omega)^(2))/(8R)`C. `((Bpiromega)^(2))/(2R)`D. `((Bpiromega^(2))^(2))/(8R)`
Answer» Correct Answer - B `phi=B(pi r^(2))/(2)cos omegat` `:. epsi=-(dphi)/(dt)=(1)/(2) B pi r^(2) omega sin omegat` `:.P=epsi^(2)/(R)=(B^(2)pi^(2)r^(4)omega^(2)sin^(2)omegat)/(4R)` `:. ltPgt=((Bpir^(2)omega)^(2))/(8R)" " ( :.ltsin^(2)omegatgt=(1)/(2))`

Discussion

288.	A circuit area `0.01m^(2)` is kept inside a magnetic field which is normal to its plane. The magentic field changes form 2 tesla in 1 millisecond. If the resistance of the circuit is `2omega`. The rate of heat evolved isA. `5J//s`B. `50J//s`C. `0.05J//s`D. `0.5J//s`
Answer» Correct Answer - B Induced emf `,e=(-dphi)/(dt)` `rArre=(-(phi_(2)-phi_(1)))/(t)=(-(0.01-0.02))/(1xx10^(-3))` `e=0.01xx10^(3)=10` volt. `V=IR or (10)/(2)=I:. I=5A` Rate of heat evolved `=I^(2)R=(5)^(2)xx2=25xx2=50J//s`

Discussion

289.	Some magnetic flux is changed from a coil resistance `10Omega`. As a result an induced current developed in it. Which varies with time as shown figure, The magnitude of changes f in flux through the coil (in webers) is A. 2B. 4C. 6D. None of these
Answer» Correct Answer - A Charge induced in coil is given as `dq=(dphi_(B))/(R)` = I dt = Area under `i-t` graph `therefore" "dphi_(B)` =(Area under `i-t` graph)R `=(1)/(2)xx4xx0.1xx10=2Wb`

Discussion

290.	A coil has an area of `0.05 m^(2)` and it has `800` turns. It is placed perpendicular in a magnitude field of strength `4xx10^(-5)Wb//m^(2)`, it is rotated through `90^(@)` in `0.1` sec. the average e.m.f. induced in the coil isA. 0.056 VB. 0.046 VC. 0.026 VD. 0.016 V
Answer» Correct Answer - D Induced emf, `e=\|(-NBA(costheta_(2)-costheta_(1)))/(Deltat)\|` `=\|-(800xx4xx10^(-5)xx0.05(cos90^(@)-cos0^(@)))/(0.1)\|=0.016V`

Discussion

291.	A circular coil of radius 6 cm and 20 turns rotates about its vertical diameter with an angular speed of `40"rad"s^(-1)` in a uniform horizontal magnetic field of magnitude `2xx10^(-2)T`. If the coil form a closed loop of resistance `8omega`, then the average power loss due to joule heating isA. `2.07xx10^(-3)W`B. `1.23xx10^(-3)W`C. `3.14xx10^(-3)W`D. `1.80xx10^(-3)W`
Answer» Correct Answer - A Here, `r=6cm =6xx10^(-2)m`, `N=20,omega=40 "rad" s^(-1)` `B=2xx10^(-2)T,R=8Omega` Maximum emf induced, `epsi=NABomega` `=N(pir^(2))Bomega` `=20xxpixx(6xx10^(-2))^(2)xx2xx10^(-2)xx40=0.18V` Average value of emf induced over a full cycle `epsi_(av)-0` Mixmum, value of current in the coil, `I=(epsi)/(R)=(0.18)/(8)=0.023A` Average power dissipated, `P=(epsil)/(2)=(0.18xx0.023)/(2)=2.07xx10^)-3) W`

Discussion

292.	A rectangular coil of `300` turns has an average area of average area of `25 cmxx10 cm` the cooil rotates with a speed of `50 cps ` in a uniform magnetic field of strength `4xx10^(-2)T` about an axis perpendicular of the field. The peak value of the induced e.m.f. is (in volt)`A. `3000pi`B. `300pi`C. `30pi`D. `3pi`
Answer» Correct Answer - C Peak value of emf `=e_(0)=omegaNBA=2pivNBA` `=2pixx50xx300xx4xx10^(-2)xx(25xx10^(-2)xx10^(-2))=30pi` volt

Discussion

293.	A circular coil of radius `5` cm has `500` turns of a wire. The approximate value of the coefficient of self-induction of the coil will beA. `25 millihenry`B. `25xx10^(-3)` millihenryC. `50xx10^(-3)` millihenryD. `50xx10^(-3)` henry
Answer» Correct Answer - A `varphi=LiimpliesNBA=LI` since magnetic field at the centre of circular coil carrying current is given by `B=(mu_(0))/(4pi).(2piNi)/(R )` `:. N.(mu_(0))/(4pi).(2piNi)/(R).pir^(2)=LiimpliesL(mu_(0)N^(2)pir)/(2)` Hence self-inductance of a coil `=(4pixx10^(-7)xx500xx500xxpixx0.05)/(2)=25 mH`

Discussion

294.	The total charge induced in a conducting loop when it is moved in magnetic field depends onA. The rate of change of magnetic fluxB. initial magnetic flux onlyC. the total change in magnetic fluxD. final magnetic flux only
Answer» Correct Answer - C `q=(N)/(R )dvarphi:.qpropvarphi`

Discussion

295.	A coil of `40 Omega` resistance has `100` turns and radius `6 mm` us connected to ammeter of resistance of `160 ohms`. Coil is placed perpendicular to the magnetic field. When coil is taken out of the field, `32 mu C` charge flows through it. The intensity of magnetic field will beA. 6.55 TB. 5.66 TC. 2.55 TD. 0.566 T
Answer» Correct Answer - D We have, `Deltaq=-(N)/(R)Deltaphi and Deltaphi=DeltaB.A cos theta` `therefore" "32xx10^(-6)=-(100)/(160+40)(0-B)xxpixx(6xx10^(-3))^(2)xxcos0^(@)` `therefore` Intensity of magnetic field, `B=0.566T`

Discussion

296.	A coil of `40 Omega` resistance has `100` turns and radius `6 mm` us connected to ammeter of resistance of `160 ohms`. Coil is placed perpendicular to the magnetic field. When coil is taken out of the field, `32 mu C` charge flows through it. The intensity of magnetic field will beA. 0.665 TB. 0.656 TC. 0.566 TD. 5.666 T
Answer» Correct Answer - C `phi=(phi_(1)-phi_(2))/(R)=(nAB)/(R)` `therefore B=(phiR)/(nA)=0.566T`

Discussion

297.	How much charge will flow through a `200 Omega` galvanometer connected to a `400 Omega` circular coil of 1000 turns wound on a wooden stick 20 mm in diameter, if a magnetic field b = 0.0113 T parallel to the axis of the stick is decreased suddenly to zero.
Answer» Here, total resistance of the circuit, `R = 200 + 400 = 600 Omega` `n = 1000, D = 20 mm, r = 10 mm = 10^(-2) m`. `:.` Area of cross-section `A = pi r^(2) = pi (10^(-2))^(2) m^(2) = 10^(-4) pi m^(2)` `B_(1) = 0.0113 T, B_(2) = 0, :. Db = B_(2) - B_(1) = - 0.0113 T` Now `e = IR = (dq)/(dt) R` Also, `e = - (n d phi)/(dt) = (-n A dB)/(dt)` From (i) and (ii), `dq = (-n A dB)/(R ) = (- 1000 xx 10^(-4) pi xx (-0.0113))/(600) = + 5.9 xx 10^(-6) C`

Discussion

298.	A closed coil consists of 500 turns has area `4 cm^2` and a resistance of `50 Omega`. The coil is kept with its plane perpendicular to a uniform magnetic field of `0.2 Wb/m^2`. Calculate the amount charge flowing through the coil if it is rotated through `180^@`
Answer» Correct Answer - `1.6xx10^(-3)` C,No

Discussion

299.	A circular coil opf `500` turns of wire has an enclosed area of `0.1 m^(2)` per turn. It is kept perpendicular to a magnetic field of induction `0.2 T` and rotated by `180^(@)` about a diameter perpendicular to the field in `0.1` sec. how much charge will pass when the coil is connected to a gavanometer with a combined resistance of `50 ohms`A. `0.2 C`B. `0.4 C`C. `2 C`D. `4 C`
Answer» Correct Answer - B `DeltaQ=(NBA)/(R )(costheta_(1)-costheta_(2))` `=(500xx0.2xx0.1(cos0-cos180))/(50)=0.4C`

Discussion

300.	A coil of `40 Omega` resistance has `100` turns and radius `6 mm` us connected to ammeter of resistance of `160 ohms`. Coil is placed perpendicular to the magnetic field. When coil is taken out of the field, `32 mu C` charge flows through it. The intensity of magnetic field will beA. `6.55 T`B. `5.66 T`C. `0.655 T`D. `0.566 T``
Answer» Correct Answer - D `q=(N)/(R )(B_(2)-B_(1))A costheta` `32xx10^(-4)` `=-(100)/((160+40))(0-B)xxpixx(6xx10^(-3))^(2)xxcos0^(@)` `impliesB=0.565T`

Discussion

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