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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.
| 351. |
The deflection in a moving coil galvanometer isA. Directly proportional to the torsional constantB. Directly proportional to the number of turns in the coilC. Inversely proportional to the area of the coilD. Inversely proportional to the current flowing |
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Answer» Correct Answer - B |
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| 352. |
IF the current is doubled, the deflection is also doubled inA. A tangent galvanometerB. A moving coil galvanometerC. Both (a) and (b)D. None of these |
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Answer» Correct Answer - B |
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| 353. |
Which is a vector quantityA. DensityB. Magnetic fluxC. Intensity of magnetic fieldD. Magnetic potential |
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Answer» Correct Answer - C |
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| 354. |
A resistance of `900Omega` is connected in series with a galvanometer of resistance `100Omega`. A potential difference of 1 Volt produces 100 division deflection in galvanometer. Find the figure of merit of galvanometer. |
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Answer» Correct Answer - `10^-5A//div` Figure of merits, `K=1/theta=(V)/(Rtheta)=(1)/((900+100)100)` `=10^-5A//d iv`. |
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| 355. |
A particle of charge `16xx10^(-18)` coulomb moving with velocity `10m//s` along the `x-` axis enters a region where a magnetic field of induction B is along the `y-` axis, and an electric field of magnitude `10//m^(-1)` is along the negative `Z-` axis. If the charged particle continues moving along the `X-` axis, the magnitude to B isA. `10^(-3)Wb//m^(2)`B. `10^(3)Wb//m^(2)`C. `10^(5)Wb//m^(2)`D. `10^(16)Wb//m^(2)` |
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Answer» Correct Answer - B |
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| 356. |
Two coaxil solenoids 1 and 2 of the same length are set so that one is inside the other. The number of turns per unit length are `n_(1)` and `n_(2)`. The current `i_(1)` and `i_(2)` are flowing in opposite directions. The magnetic field inside the inner coil is zero. This is possible whenA. `i_(1)!=i_(2)` and `n_(1)=n_(2)`B. `i_(1)=i_(2)` and `n_(1)!=n_(2)`C. `i_(1)=i_(2)` and `n_(1)=n_(2)`D. `i_(1)n_(1)=i_(2)n_(2)` |
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Answer» Correct Answer - C::D |
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| 357. |
Which of the follwing particles will have minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field?A. `Li`B. ElectronC. ProtonD. `He^(+)` |
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Answer» Correct Answer - A |
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| 358. |
An element `Deltavecl=Deltaxhati` is placed at the origin as shown in figure and carries a current `I=2A`. Find out the magnetic field at a point P on the y-axis at a distance of `1*0m` due to the element `Deltax=1cm`. Give also the direction of the field produced. |
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Answer» Correct Answer - `2xx10^-9T` in +z-direciton Here, `Idvecl=IDeltaxhati`, where `I=2A`, `Deltax=1cm=10^-2m`, `r=1*0m` and `vecr=1*0hatj` `dvecB=(mu_0)/(4pi)(Idveclxxvecr)/(r^3)=(mu_0)/(4pi)(IDeltaxhatixx(1*0hatj))/(r^3)` `=(mu_0)/(4pi)((IDeltaxxx1*0))/(r^3)(hatixxhatj)=(mu_0)/(4pi)(IDeltaxxx1*0)/(r^3)xxhatk` `=10^-7xx(2xx10^-2xx1*0)/(1^3)hatk` `=2xx10^-9T` in +z direction. |
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| 359. |
An element, `Deltavect=Deltaxhati` is placed at the origin as shown in figure and carries a current 5A. Find out the magnetic field at point P on the z-axis at a distance of `2-0m` due to the element `Deltax=1cm`. Give also, the direction of the magnetic field produced. |
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Answer» Here, `Idvecl=I Deltaxhati`, where `I=5A` `Deltax=1cm=10^-2m, r=2.0m, vecr=2.0hatk` `dvecB=(mu_0)/(4pi)(Idveclxxvecr)/(r^3)=(mu_0)/(4pi)(I(Deltaxhati)xx(2.0hatk))/(r^3)` `10^-7xx(5xx10^-2xx2.0)/(2^3)(-hatj)` `=1.25xx10^-9T` along `-y` direction |
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| 360. |
A solenoid of `500turns` per meter is carrying a current of `3A`. Its core is made of iron, which has a relative permeability of `5000`. Determine the magnitudes of magnetic intensity, intensity of magnetization and magnetic field inside the core. |
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Answer» Here, `n = 500` turns/m, `i = 3A, mu_(r) = 5000` Magnetic intensity, `H = n i = 500 m^(-1) xx 3A = 1500 Am^(-1)` As `mu_(r) = 1 + chi_(m)` `:. Chi_(m) = mu_(r) - 1 = 5000 - 1 = 4999 ~~ 5000` Also, `mu_(r) =(mu)/(mu_(0)) = 5000 :. mu = 5000 mu_(0)` Intensity of magnetisation, `I = chi_(m) H = 5000 xx 1500 = 7.5 xx 10^(6) Am^(-1)` Magnetic field inside the core, `B = mu H = 5000 mu_(0)H` `= 5000 (4pi xx 10^(-7)) xx 1500` `B = 3 pi = 3 xx (22)/(7) = 9.4T` |
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| 361. |
Can we decrease or increase the range of the given voltmeter? |
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Answer» We can increase the range of the given voltmeter by putting a suitable high resistance in series with the voltmeter. We can decrease the range of the given voltmeter by putting a suitable resistance in parallel with the high resistance already in series with the galvanometer working as voltmeter, so that the effective resistance of the voltmeter multiplied with the current `(I_g)` gives the required potential difference to be measured now. |
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| 362. |
A voltmeter, an ammeter and a resistance are connected in series with a lead accumulator. The voltmeter gives some deflection but the deflection of ammeter is almost zero. Explain why? |
| Answer» Voltmeter resistance being very high, when connected in series, it makes the effective resistance of the circuit very high. Due to it, the current in the circuit becomes extremely small. Since ammeter measures the current, hence the deflection of ammeter is almost zero. As voltmeter measures potential difference between the two points, it will show the reading due to voltage of the lead accumulator. | |
| 363. |
While convertying galvanometer into an ammeter, how does the parallel low resistance (shunt) bring the required changes in the galvanometer? |
| Answer» When a galvanometer is to be converted into an ammeter of a given range, a suitable low resistance `S=I_gG(I-I_g)` is connected in parallel with the galvanometer. In doing so, the limited current of circuit flows through galvanometer and remaining current of the circuit flows through shunt. Due to it, a galvanometer works as an ammeter of given range. | |
| 364. |
Why do we not use galvanometer as an ammeter? |
| Answer» A galvanometer shows a full scale deflection with a very small current. Hence a galvanometer can measure limited current. Therefore as such, a galvanometer can not be used as an ammeter, which can measure the given large current. | |
| 365. |
Why should a voltmeter have a low current carrying capacity? |
| Answer» A voltmeter should have a low current carrying capacity so that its resistance is high. It draws a very small current from the circuit. Due to it, the potential difference `(V=IR)` to be measured will not be much different from the actual value. | |
| 366. |
Why should an ammeter have a low resistance and a high current carrying capacity? |
| Answer» An ammeter is used to measure the current.It can measure the current of the circuit if connected in series of the circuit. The ammeter connected in series of the circuit. The ammeter connected in series of the circuit can measure the current and will not disturb the current of the circuit if its resistance is low and current carrying capacity is high. | |
| 367. |
Why should an ammeter have a high current carrying capacity? |
| Answer» An ammeter should have a high current carrying capacity so that its resistance is least. Its presence in the circuit does not affect the current in the circuit to be measured. | |
| 368. |
What is the difference between solenoid and toroid? |
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Answer» A solenoid consists of an insulating long wire closely wound in the form of helix. Its length is very large as compared to its diameter. A toroid is a hollow circular ring on which a large number of insulated turns of a metallic wire are closely wound. Infact, toroid is an endless solenoid. |
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| 369. |
What is the magnitude and direction of force on an electron moving along the direction of the magnetic field. |
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Answer» When an electron is moving along the direction of the magnetic field, then angle `theta` between velocity vector `vecv` and magnetic field `vecB` is zero degree, i.e., `theta=0^@`. Lorentz magnetic force `F=-evBsin0^@=0`. |
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| 370. |
Is the source of magnetic field analogue to the source of electric current. |
| Answer» No, because the source of magnetic field is a magnetic dipole, but the source of electric field is an electric charge. | |
| 371. |
A beam of electrons starts to accelerate from rest due to a uniform electric field in vacuum, the moving electrons:A. initially experience a force of mtual repulsionB. experience a force of mutual attraction after travelling a certain distanceC. will continue to diverge due to electrostatic repulsionD. will follow parallel lines because there is no force of attraction between them. |
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Answer» Correct Answer - A::B The beam of electrons consists of number of electrons. In moving beam of electrons two types of forces play their part. (i) Electrostatic respulsion due to like charges on electrons and (ii) Magnetic attraction, as the two moving electrons in the same direction behave as two parallel linear conductors carrying current in the same directions. Initially electrostatic repulsion is more effective than the magnetic attraction as speed of electrons is small. When speed of electrons become high, the magnetic attraction becomes more prominent than electrostatic repulsion. |
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| 372. |
An electron beam moving with uniform velocity is gradually diverging. When it is accelerated to a very high velocity, it again starts converging. Why? |
| Answer» Moving electrons, apar from electrical repulsion experience magnetic attraction also. If the electron beam is moving under normal conditions, the electrical repulsive force is much stronger than the magnetic attraction and hence the beam diverges. When the electron beam is moving at very high velocity, the magnetic force of attraction becomes more effective than electrical repulsion and the beam starts converging. | |
| 373. |
A thin disc of dielectric material, with a total charge +q distributed uniformly over its surface rotates n times per second about an axis perpendicualr to the surface of the disc and passing through its center. Find the magnetic induction at the center of the disc. The radius of the disc is equal to r. |
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Answer» Correct Answer - `(mu_(0)nq)/(r )` |
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| 374. |
A thin circular disk of radius `R` is uniformly charged with density `sigma gt 0` per unit area.The disk rotates about its axis with a uniform angular speed `omega`.The magnetic moment of the disk is :A. `pi R^(2) a omega`B. `(piR^(4))/(4) alpha omega`C. `(piR^(4))/(4) alpha omega`D. `2pi R^(4) alpha omega` |
| Answer» Correct Answer - 1 | |
| 375. |
Two parallel wires are carrying electric currents of equal magnitude and in the same direction. They excertA. An attractive force on each otherB. A repulsive force on each otherC. No force on each otherD. A rotational torque on each other |
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Answer» Correct Answer - A |
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| 376. |
Which has greater resistance (a) milliammeter or ammeter? (b) milli voltmeter or voltmeter? |
| Answer» The resistance of milliammeter is greater than that of ammeter. The resistance of voltmeter is greater than that of millivoltmeter. | |
| 377. |
If the bar magnet in the above problem is turned around by `180^@`, where will the new null points be located? |
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Answer» When the bar magnet is turned through `180^@`, neutral points would lie on equatorial line, so that `B_2=(mu_0)/(4pi)(M)/(d_2^3)=H` …(i) In the previous question, `B_1=(mu_0)/(4pi)(2M)/(d_1^3)=H` …(ii) From (i) and (ii), `(mu_0)/(4pi)(M)/(d_2^3)=(mu_0)/(4pi)(2M)/(d_1^3) :. d_2^3=d_1^3/2=((14)^3)/(2)` `d_2=(14)/(2^(1//3)=11*1cm` |
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| 378. |
A straight wire of diametre 0.5 mm carrying a current of 1 A is replaced by another wire of 1mm diametre carrying same current. The strenth of magnetic field far away isA. twice the earlier valueB. same as the earlier valueC. one-half of the earlier valueD. one-quarter of the earlier value |
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Answer» Correct Answer - b |
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| 379. |
An electron travels in a circular path of radius `20cm` in a magnetic field `2xx10^-3T` (i) Calculate the speed of the electron (ii) What is the potential difference through which the electron must be accelerated to acquire this speed? Charge of electron `=1*6xx10^(-19)C`. Mass of electron `=9xx10^(-31)kg`. |
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Answer» Here `r=20xx10^-2m`, `B=2xx10^-3T`, (i) Magnetic force on the electron=centripetal force or `evB=(mv^2)/(r)` or `v=(eBr)/(m)` `=((1*6xx10^(-19))xx(2xx10^-3)xx(20xx10^2))/(9xx10^(-31))` `=7*1xx10^7ms^-1` (ii) Let V be the pot. diff. required to provide speed v to the electron, then `eV=1/2mv^2` or `V=(mv^2)/(2e)=(9xx10^(-31)xx(7*1xx10^7)^2)/(2xx(1*6xx10^(-19))` `=14*2xx10^3V=14*2kV` |
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| 380. |
A bar magnetic made of steel has a magnetic moment of `2*5Am^2` and a mass of `6*6g`. If density of stell is `7*9xx10^-3kg//m^3`, what is the intensity of magnetisation? |
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Answer» Here, `M=2*5Am^2, m=6*6g=6*6xx10^-3kg` `rho=7*9xx10^3kg//m^3, I=?` `I=M/V=(M)/(m//rho)=(Mrho)/(m)=(2*5xx7*9xx10^3)/(6*6xx10^-3)=2*99xx10^6Am^-1` |
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| 381. |
What is the basic difference between magnetic and electric lines of force? |
| Answer» Magnetic lines of force are closed, continuous curves, but electric lines of force are continuous loops which are not closed. | |
| 382. |
Give two examples of magnetic dipole. |
| Answer» Every atom of para and ferro magnetic substances, a loop of current are magnetic dipoles. | |
| 383. |
On what factors does the pole strength of a magnet depend? |
| Answer» The pole strengths of a magnet may depend on its cross section, nature and state of magnetisation. | |
| 384. |
A beam of alpha particle passes underflected with a horizontal velocity v, through a region of electric and magnetic fields, mutually perpendicular to each other and normal to the direction of the beam. If the mangnitude of the electric and magnetic fields are `120kVm^-1` and `60mT` respectively, calculate (i) velocity of the beam (ii) force with which it strikes a target on a screen, if the alpha particle beam current is equal to `1mA`? Mass of proton=mass of neutron `=1*675xx10^(-27)kg`. |
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Answer» Here, `E=120kVm^-1` `=120xx10^3Vm^-1`, `B=60mT=60xx10^-3T` (i) For undeflected beam of alpha particle `qvB=qE` or `v=E/B=(120xx10^3)/(60xx10^-3)` `=2xx10^6ms^-1` (ii) Current carried by alpha particle beam `I=1mA=10^-3A` Number of alpha particles striking the screen per second `n=I/q=(I)/(2e)=(10^-3)/(2xx1*6xx10^(-19))` `=3*125xx10^(15)s^-1` Mass of alpha particle, `m_alpha=2m_p+2m_n=4m_p` `=4xx1*675xx10^(-27)kg` Force with which alpha particle beam strikes a target on the screen `F=m_alphanv=(4xx1*675xx10^(-27))` `xx(3*125xx10^(15))xx2xx10^6` `=4*19xx10^-5N` |
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| 385. |
Magnetization and demagnetization of soft iron is easier/more difficult as compared steel. Why? |
| Answer» Magnetisation and demagnetisation of soft iron is easier compared to steel, because coercivity, of soft iron is smaller than the coercivity of steel. | |
| 386. |
To make the field radial in a moving coil galvanometerA. The number of turns in the coil is increasedB. Magnet is taken in the form of horse-shoeC. Poles are cylindrically cutD. Coil is wounded on aluminium frame |
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Answer» Correct Answer - C |
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| 387. |
If a wire of length `l` meter placed in uniform magnetic field 1.5 Tesla at angle `30^(@)` with magnetic field. The current in a wire 10 amp. Then force on a wire will beA. `7.5N`B. `1.5N`C. `0.5N`D. `2.5N` |
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Answer» Correct Answer - A |
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| 388. |
What is the shape of magnet in moving coil galvanometer to make the radial magnetic field ?A. ConcaveB. Horse shoe magnetC. ConvexD. None of these |
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Answer» Correct Answer - A |
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| 389. |
Two identical charged particles enter a uniform magnetic field with same speed but at angles `30^(@)` and `60^(@)` with field Let a,b and c be the ratio of their time periods, radii and pitches of the helical paths than .A. `abc =1`B. `abcgt1`C. `abc lt 1`D. `a =bc` |
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Answer» Correct Answer - A,D `r =T = (2 pim)/(qB) a =1` `r =(mV_(1))/(qB), b = (sin 30^(@))/(sin 60^(@)) = (1)/(sqrt3)` Pitch a `v cos0 c = (cos 30^(@))/(cos 60^(@)` `So abc =1` . |
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| 390. |
Which of the following statement is correct .A. A charged particle enters a region of uniform magnetic field at an angle `85^(@)` to magnetic lines of force. The path of the particle is a circle .B. An electron and proton are moving with the same kinetic energy along the same direction. When they pass through uniform magnetic field perpendicular to their direction of motion, they describe circular path.C. There is no change in the energy of a charged particle moving in a magnetic field although magnetic force acts on it .D. Two electrons enter with the same speed but in opposite direction in a magnetic field. Then the two describe circle of the same radius and these move in the same direction . |
| Answer» Correct Answer - B,C,D | |
| 391. |
In the previous question, if the current is I and the magnetic field at `D` has magnitude `B` .A. `B=(mu_(0)i)/(2sqrt2pi)`B. `B=(mu_(0)i)/(2sqrt23pi)`C. `B` is parallel to the x-axisD. `B` makes an angle of `45^(@)` with the xy plane |
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Answer» Correct Answer - A,D `B =(mu_(0))/(4pi) (2I)/(sqrt2) = (4 piI)/(2sqrt2 pi` It will have not any component along x -axis any point . |
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| 392. |
A straight rod of mass m and lenth `L` is suspended from the identical spring as shwon in the figure The spring strectched by a distance of `x_(0)` due to the weight of the wire The circuit has total resistance `Romega` When the magnetic field perpendicular to the plane of the paper is switched on, springs are observed to extend further by the same distance The magnetic field strenght is A. `2(mgR)/(LV)`B. `(mgR)/(LV)`C. `(mgR)/(2LV)`D. `(mgR)/(V)` |
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Answer» Correct Answer - A `mg = kx …(1) `mg + ILB = k. 2x…(2)` From(I) and (2)…(2)` `ILB = mg implies B =(mg)/(IL) o.` `B = (mgR)/(EL)o.` . |
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| 393. |
The direction of magnetic force on the electron as shown in the diagram is along .A. `y-"axis"`B. `-y-"axis"`C. `z-"axis"`D. `-z-"axis"` |
| Answer» Correct Answer - A | |
| 394. |
A block of mass m & charge q is released on a long smooth inclined plane magnetic field `B` is constant, unifrom, horizontal and parallel to surface a shown Find the time from start when block loses contact with the surface .A. `(m costheta)/(qB)`B. `(mcosectheta)/(qB)`C. `(mcottheta)/(qB)`D. none |
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Answer» Correct Answer - C Block will loses confact when force due to magnetic field will become equal to mag costheta `qVB = mg cos theta` `V = (mg costheta)/(qB) = g sinthetat = (m cottheta)/(qB)` . |
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| 395. |
A hollow cylinder having infinite length and carrying unifrom current per unit length `lamda` along the circumference as shown Magnetic field inside the cylinder is .A. `(mu_(0)lamda)/(2)`B. `mu_(0)lambda`C. `2mu_(0)lambda`D. none |
| Answer» Correct Answer - B | |
| 396. |
Why are pole pieces of galvanometer made concave? |
| Answer» To have a uniform, strong and radial magnetic field. | |
| 397. |
Why is the coil wrapped on a conducting frame in a galvanometer? |
| Answer» When a coil is rotated in the magnetic field, a variable magnetic flux is linked with the conducting frame of the coil, resulting the eddy currens in it, which opposes the cause producing it. These eddy currents in conducting frame help in stopping the coil soon i.e., in making the galvanometer dead beat. | |
| 398. |
What is the function of soft iron cylinder between the poles of a galvanmeter? |
| Answer» It concentrates the magnetic field and helps in making the magnetic field radial and strong. | |
| 399. |
A particle of charge q and mass `m` is moving along the x-axis with a velocity `v` and enters a region of electric field `E` and magnetic field `B` as shown in figures below. For which figure the net force on the charge may be zero?A. B. C. D. |
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Answer» Correct Answer - B |
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| 400. |
The correct curve between the magnetic induction (B) along the axis of a along solenoid due to current flow i in it and distance x from one end is -A. B. C. D. |
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Answer» Correct Answer - A |
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