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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.
| 251. |
A long wire `A` carries a current of `10` amp. Another long wire `B`, which is parallel to `A` and separated by `0.1m` from `A`, carries a current of `5` amp, in the opposite direction to that in `A`. What is the magnitude and nature of the force experienced per unit length of `B`? `(mu_(0)=4pixx10^(-7)weber/amp-m)`A. Repulsive force of `10^(-4)N//m`B. Attractive force of `10^(-4)N//m`C. Repulsive force of `2pixx10^(-5)N//m`D. Attractive force of `2pixx10^(-5)N//m` |
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Answer» Correct Answer - A |
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| 252. |
A partical having charge of `1.C` , mass `1kg` and speed `1m//s` enters a uniform magnetic field having magnetic induction of `1T` at an angle `0=30^(@)` between velocity vector and magnetic induction The pitch of its helical path is (in meters) .A. `(sqrt3pi)/(2)`B. `sqrt3pi`C. `(pi)/(2)`D. `pi` |
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Answer» Correct Answer - B Pitch ` = V cos theta xx T = Vcostheta xx (2pim)/(qB) = sqrt3 pi` . |
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| 253. |
Two long parallel wires carrying equal current separated by `1m`, exert a force of `2xx10^(-7)N//m on one another. The current flowing through them isA. `2.0A`B. `2.0xx10^(-7)A`C. `1.0A`D. `1.0xx10^(-7)A` |
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Answer» Correct Answer - C |
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| 254. |
Two long and parallel wires are at a distance of `0.1m` and a current of `5 A` is flowing in each of these wires. The force per unit length due to these wires will beA. `5xx10^(-5)N//m`B. `5xx10^(-3)N//m`C. `2.5xx10^(-5)N//m`D. `2.5xx10^(-4)N//m` |
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Answer» Correct Answer - A |
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| 255. |
What is the expression for potential energy of a dipole of moment M held at an angle `theta` with a magnetic field B? |
| Answer» Correct Answer - `P.E.=-MBcostheta` | |
| 256. |
A charged particle (charge `q`) is moving in a circle of radius `R` with unifrom speed `v`. The associated magnetic moment `mu` is given byA. `(qvR)/(2)`B. `qvR^(2)`C. `(qvR_(2))/(2)`D. qvR |
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Answer» Correct Answer - a |
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| 257. |
A circular loop of radius 0.3 cm lies parallel to amuch bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop isA. `3.3 xx 10^(11)` weberB. `6.6 xx 10^(-9)` weberC. `9.1 xx 10^(11)` weaberD. `6 xx 10^(-11)` weber |
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Answer» Correct Answer - 3 `phi = (mu_(0)lR^(2))/(2 (R^(2) + x^(2))^(3//2))xx pi r^(2)` `= (mu_(0)lR^(2)r^(2)pi)/(2(R^(2 +x^(2)))^(3//2))` `= (4 pi xx 10^(-7 xx 2 xx (0.2)^(2) (0.03)^(2) xx pi))/(2[(0.2)^(2) + (0.15)^(2)]^(3//2))` ` = 9.09 xx 10^(11)` . |
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| 258. |
Under the influence of a unifrom magnetic field a charged particle is moving on a circle of radius `R` with Constnant speed `v`. The time period of the motionA. depends on v and not on RB. Depends on both R and vC. is independent of both R and vD. depends on R and not on v |
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Answer» Correct Answer - c |
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| 259. |
The magntic moment `(mu)` of a revolving electron around the mucleaus varies with principle quantum number `n` asA. both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.B. both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both Assertion and Reason are false. |
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Answer» Correct Answer - D Magnetic moment of revolving electron around the nucleus is, `M=IA=e/Tpir^2=(eomega)/(2pi)pir^2=(evr)/(2)` But `vprop1/n` and `rpropn^2` Therefore, `Mprop(e(1//n)xxn^2)/(2)` or `Mpropn` The Assertion is wrong. Here Reason though correct but cannot explain the Assertion. |
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| 260. |
Two long parallel wires `P` and `Q` are both perpendicular to the plane of the paper with distance `5m` between them. If `P` and `Q` carry current of `2.5` amp and `5` amp respectively in the same direction, then the magnetic field at a point half way between the wires isA. `(sqrt(3)mu_(0))/(2pi)`B. `(mu_(0))/(pi)`C. `(3mu_(0))/(2pi)`D. `(mu_(0))/(2pi)` |
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Answer» Correct Answer - D |
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| 261. |
What will be the resulatant field at the origin due to the four indinite length wires if each wire produces magnetic field `B` at origin ? A. `4B`B. `sqrt(2)B`C. `2sqrt(2)B`D. Zero |
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Answer» Correct Answer - C |
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| 262. |
A straight wire of length `(2pi^(2))` metre is carrying a current of `2A` and the magnetic field due to it is measured at a point distant `1 cm` from it. If the wire is to be bent into a circle and is to carry the same current as before, the ratio of the magnetic field at its centre to that obtained in the first case would beA. `50:1`B. `1:50`C. `100:1`D. `1:100` |
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Answer» Correct Answer - B |
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| 263. |
A long copper tube of inner radius `R` carriers a current `i`. The magnetic field `B` inside the tube isA. `(mu_(0)i)/(2piR)`B. `(mu_(0)i)/(4piR)`C. `(mu_(0)i)/(2R)`D. Zero |
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Answer» Correct Answer - D |
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| 264. |
A current of one ampere is passed through a straight wire of length `2*0` metre. Find the magnetic field at a point in air at a distance 3 metre from one end of wire but lying on the axis of the wire.A. `(mu_(0))/(2pi)`B. `(mu_(0))/(4pi)`C. `(mu_(0))/(8pi)`D. Zero |
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Answer» Correct Answer - D |
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| 265. |
Mixed `He^(+)` and `O^(2+)` ions (mass of `He^(+)`=4 amu and that of `O^(2+)=16` amu) beam passes a region of constant perpendicular magnetic field. If kinetic energy of all the ions is same thenA. `He^(+)` ions will be deflected more than tose of `O^(2+)`B. `He^(+)` ions will be deflected less than those of `O^(2+)`C. All the ions will be deflected equallyD. No ions will be deflected |
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Answer» Correct Answer - C |
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| 266. |
The magnetic field at the centre of a circular coil of radius `r` carrying current `l` is `B_(1)`. The field at the centre of another coil of radius `2r` carrying same current `l` is `B_(2)`. The ratio `(B_(1))/(B_(2))` isA. `1/2`B. 1C. 2D. 4 |
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Answer» Correct Answer - C |
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| 267. |
A battery is connected between two points `A and B` on the circumference of a uniform conducting ring of radius `r` and resistance `R` . One of the arcs `AB` of the ring subtends an angle `theta` at the centre . The value of the magnetic induction at the centre due to the current in the ring isA. Proportional to `2(180^(@)-theta)`B. Inversely proportinal to `r`C. Zero, only if `theta=180^(@)`D. Zero for all values of `theta` |
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Answer» Correct Answer - D |
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| 268. |
A cell is connected between the point A and C of a circular conductor ABCD of centre `O, /_AOC = 60^(@)`. If `B_(1_` and `B_(2)` are the magnitude of magnetic fields at O due to the currents in ABC and ADC respectively, the ratio of `B_(1)//B_(2)` is. A. `0.2`B. `6`C. `1`D. `5` |
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Answer» Correct Answer - C |
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| 269. |
A current of `0.1 A` circulates around a coil of 100 turns and having a radius equal to `5cm`. The magnetic field set up at the centre of the coil is `(mu=4pixx10^(-5) "weber"//"amp-metre")`A. `2xx10^(-5)` TeslaB. `4xx10^(-5)` TeslaC. `8pixx10^(-5)` TeslaD. `4pixx10^(-5)` Tesla |
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Answer» Correct Answer - D |
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| 270. |
A current carrying long solenoid is placed on the ground with its axis vertical. A proton is falling along the axis of the solenoid with a velocity `v`. When the proton enters into the solenoid, it willA. Be deflected from its pathB. Be accelerated along the same pathC. Be decelerated along the same pathD. Move along the same path with no change in velocity |
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Answer» Correct Answer - D |
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| 271. |
State two methods to destroy the magnetism of a magnet. |
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Answer» (i) By heating the magnet. (ii) By applying magnetic field in the reverse direction. |
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| 272. |
A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B. What is the magnitude of the magnetic field? A. `2`B. `1*5`C. `0*55`D. `0*65` |
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Answer» Correct Answer - D Here, `m=200g=200xx10^-3kg`, `l=1*5m`, `I=2A`, `g=9*8m//s^2`, `theta=90^@` Force on current carrying conductor in mag. field is given by `F=IlB sin theta=IlBsin 90^@=IlB` Weight `W=mg=200xx10^-3xx9*8N` As it is suspended in mid air, So `W=F` `mg=IlB` or `B=(mg)/(Il)=(200xx10^-3xx9*8)/(2xx1*5)=0*65T` |
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| 273. |
A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B. What is the magnitude of the magnetic field? |
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Answer» Here, `m=200g=0*200kg`, `I=2A, l=1*5m`. For mid air suspension, `mg=BIl` or `B=(mg)/(Il)=(0*200xx9*8)/(2xx1*5)=0*65T` |
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| 274. |
When will a magnet in an external magnetic field be in unstable equilibrium? |
| Answer» The magnet is in unstable equilibrium, when `vecM` and `vecB` are directed opposite to each other i.e., when `theta=180^@`. Torque `=MB sin 180^@=zero` and `P.E. =-MBcos180^@=MB=maxim um`. | |
| 275. |
There are two current carrying planar coils made each from identical wires of length L. `C_1` is the circular (radius R) and `C_2` is square (side a). They are so constructed that they have same frequency of oscillation when they are placed in the same uniform `vecB` and carry the same current i. Find a in terms of R. |
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Answer» For circular coil `C_1`, `n_1=(L)/(2piR)`, Magnetic moment, `M_1=n_1iA_1=(L)/(2piR)xxixxpiR^2=(LiR)/(2)` For square coil `C_2`, `n_2=(L)/(4a)`, Magnetic moment, `M_2=n_2iA_2=(L)/(4a)xxixxa^2=(Lia)/(4)` Moment of inertia of circular coil about the diameter as axis, `I_1=(massxx (radius)^2)/(2)=(mR^2)/(2)` Moment of inertia of square coil about an axis passing through its centre parallel to breadth `I_2=(ma^2)/(12)`. Time period of oscillation of the magnet in magnetic field is given by `T=2pisqrt(I)/(MB)` and `omega=(2pi)/(T)=sqrt((MB)/(I))` `:.` `omega_1^2=(M_1B)/(I_1)` and `omega_2^2=(M_2B)/(I_2)` Given, `omega_1^2=omega_2^2`, so `(M_1)/(I_1)=(M_2)/(I_2)` or `(LIR//2)/(mR^2//2)=(LIa//4)/(ma^2//12)` On solving, `a=3R` |
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| 276. |
An infinite sheet in xy plane has a uniform surface charge density `sigma`. The thickness of the sheet is infinitesimally small. The sheet begins to move with a velocity `vec(v)=v hat(i)` (i) Find the electric field `( vec(E))` and magnetic field `(vec(B))` above and below the sheet. (ii) If the velocity of the sheet is changed to `vec(v)=vhat(k)` , find the electric and magnetic field above and below the sheet. |
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Answer» Correct Answer - (1) `E=sigma/(2in_(0))` perpendiculary away from the sheet `B=(mu_(0)sigmav)/2` parallel to the sheet in `-ve` and `+ve` y direction (2) E= same as in (1);`B=0` |
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| 277. |
An `alpha`-particle and a proton are moving in the plane of the paper in a region where there is a uniform magnetic field `vecB` directed normal to the plane of the paper. If the two particles have equal linear momenta, what will be the ratio of their trajectories in the field? |
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Answer» When a charged particle of charge q moving with velocity v is subjected to normal magnetic field B, then it describe a circular path of radius r given by `Bqv=(mv^2)/(r)` or `r=(mv)/(Bq)=(p)/(Bq)` where p=momentum of the particle For the same values of p and B, `rprop1//q` `:. r_alpha/r_p=q_p/q_alpha=(e)/(2e)=1/2` |
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| 278. |
An elastic circular wire of length `l` carries a current `I`. It is placed in a uniform magnetic field `vecB` (Out of paper) such that its plane is perpendicular to the direction of `vecB`. The wire will experience A. No forceB. A stretching forceC. A compressive forceD. A torque |
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Answer» Correct Answer - B |
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| 279. |
A particle of charge `q` and mass `m` moves in a circular orbit of radius `r` with angular speed `omega`. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends onA. `omega` and `q`B. `omega q` and `m`C. `q` and `m`D. `omega` and `m` |
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Answer» Correct Answer - C |
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| 280. |
An electron is travelling along the x-direction. It encounters a magnetic field in the y-direction. Its subsequent motion wil beA. straight line along X-directionB. a circle in the X-Z planeC. a circle in the YZ planeD. a circle in the XY plane |
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Answer» Here, `vecv=vhati` and `vecB=Bhatj`, `q=-e` As we know `vecF=q(vecvxxvecB)` so `vecF=-e(vechatixxBhatj)` `vecF=-evBhatk` [As `hatixxhatj=hatk`] So the force will be along-Z-direction, as a result of which the motion of an electron will be a circle in XZ plane. |
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| 281. |
A particle of specific charge (charge/mass) `alpha` starts moving from the origin under the action of an electric field `oversetrarrE = E_(0)hati` and magnetic field `oversetrarrB = B_(0)hatk` Its velocity at `(x_(0),y_(0).0)` is `(4hati +3hatj)` The value of `x_(0)` is .A. `(13)/(2) (alphaE_(0))/(E_(0))`B. `(16)/(2) (alphaE_(0))/(B_(0))`C. `(25)/(2alphaE_(0))`D. `(5alpha)/(2B_(0))` |
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Answer» Correct Answer - C Work done electric field `=DeltaK` `qE_(0)X_(0) = (1)/(2) mv^(2)` `x_(0) = (1)/(2) (mv^(2))/(qE_(0))` `oversetrarrV =4hati + 3hatj implies V =5 ` From (1) and (2) `x_(0) = (25)/(2 alpha E_(0))` . |
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| 282. |
In a region, steady and uniform electric and magnetic fields are present . These two fields are parallel to each other. A charged particle is released from rest in this region . The path of the particle will be aA. Straight lineB. CircleC. HelixD. Cycloid |
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Answer» Correct Answer - A |
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| 283. |
A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other . The particle will remove in aA. helix of constant pitchB. straight lineC. helix of verying pitchD. cycloid |
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Answer» Correct Answer - B Since it will have force only of electric filed so locus will be strainght line . |
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| 284. |
In a region, steady and uniform electric and magnetic fields are present . These two fields are parallel to each other. A charged particle is released from rest in this region . The path of the particle will be aA. circleB. helixC. straight lineD. ellipse |
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Answer» Correct Answer - 1 Stright line |
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| 285. |
A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity thenA. speed will decreaseB. speed will increaseC. will run towards left of direction of motionD. will turn towards right of direction of motion |
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Answer» Correct Answer - a |
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| 286. |
A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity thenA. it will turn towards right of direaction of motion .B. it will turn towards left of direction of motion .C. its velocity will decreasesD. its velocity will decreases |
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Answer» Correct Answer - 2 When electron is projected in an electric field then velocity of electron will decrease . |
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| 287. |
Which of the following substances are diamagnetic? Copper, Aluminium, Sodium, Bismuth |
| Answer» Copper and Bismuth are diamagnetic. | |
| 288. |
Two resistors `400Omega` and `800Omega` are connected in series with a `6V` battery. It is desired to measure the current in the circuit. An ammeter of `10Omega` resistance is used for this purpose figure. What will be the reading in the ammeter? Similarly, if a voltmeter of `10,000Omega` resistance is used to measure the potential difference across the `400Omega` resistor, what will be the reading in the voltmeter? |
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Answer» The circuits are shown in figure. In figure, all the resistors are in series, `:.` total resistance of the circuit, `R=400+800+10=1210Omega` Hence, current in the circuit, `I=V/R=(6)/(1210)=0*00496A` Reading in ammeter is `0*00496A` In figure, the resistors of `400Omega` and `10,000Omega` are in parallel, their effective resistance `R_p` will be `R_p=(400xx10,000)/(400+10,000)=(5000)/(13)Omega` `:.` Total resistance of the circuit `=(5000)/(13)+800=(15400)/(13)Omega` Current in the circuit, `I_1=(6)/(15400//13)=(39)/(7700)A` Potential difference across A and B `=I_1R_p=(39)/(7700)xx(5000)/(13)=(150)/(77)=1*95V` `:.` Reading of voltmeter is `1*95V` |
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| 289. |
A magnetic substance has suceptibility `-0*05` at `300K`. The magnetic susceptibility of the substance at `600K` will beA. `-0*025`B. `-0*01`C. `-0*05`D. `-0*0125` |
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Answer» Correct Answer - C As `chi_m` is negative, so the substance is diamagnetic in nature. Also `chi_m` is independent of temperature. So susceptibility at `600K` is `-0*05`. |
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| 290. |
Name the SI unit of intensity of magnetisation. |
| Answer» Correct Answer - `Am^-1`. | |
| 291. |
The radius of curvature of the path of the charged particle in a uniform magnetic field is directly proportional toA. The charge on the particleB. The momentum of the particleC. The energy of the particleD. The intensity of the field |
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Answer» Correct Answer - B |
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| 292. |
What is the SI unit of magnetic field? |
| Answer» Correct Answer - Tesla. | |
| 293. |
The radius of curvature of the path of a charged particle moving in a static uniform magnetic field isA. Directly proportional to the magnitude of the charge on the particleB. Directly proportional to the magnitude of the linear momentum of the particleC. Directly proportional to the kinetic energy of the particleD. Inversely proportional to the magnitude of the magnetic field |
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Answer» Correct Answer - B::D |
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| 294. |
A wire carrying current `I` is shaped as shown. Section `AB` is a quarter circle of radius `r`. The magnetic field is directed A. At an angle `pi//4` to the plane of the paperB. Perpendicular to the plane of the paper and directed in to the paperC. Along the bisector of the angle `ACB` towards `AB`D. Along the bisector of the angle `ACB` away from `AB` |
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Answer» Correct Answer - B |
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| 295. |
A current of 0.1A passes through a circular coil, 15 cm in diameter and consisting of 50 closely wound time of fine insulated wire. Calculate the intensity of field (H-vector) (i) at the center of the coil, (ii) at a point on its axis 7.5 cm from the center. |
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Answer» Correct Answer - `33.33 Am^(-1)`, `11.79Am^(-1)` |
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| 296. |
A charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will beA. A straight lineB. A circleC. A helix with uniform pitchD. A helix with non-uniform pitch |
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Answer» Correct Answer - C |
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| 297. |
Assertion: Above Curie temperature, a frerromagnetic material becomes paramagnetic. Reason: When a magnetic material is heated to very high temperature, it loses its magnetic properties.A. both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.B. both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both Assertion and Reason are false. |
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Answer» Correct Answer - A Here both Assertion and Reason are true and Reason is the correct explanation of Assertion because when ferromagnetic material is heated above curve temperature the domains in ferromagnetic material are broken and material behaves as a paramagnetic material. |
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| 298. |
A moving coil galvanometer has `N` numbr of turns in a coil of effective area `A`, it carries a current `I`. The magnetic field `B` is radial. The torque acting on the coil isA. `NA^(2)B^(2)I`B. `NABI^(2)`C. `N^(2)ABI`D. `NABI` |
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Answer» Correct Answer - D |
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| 299. |
A linear conductor carrying current is placed in a magnetic field. In which situation, the force experienced by the conductor is maximum and minimum? |
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Answer» The force experienced by the linear conductor carrying placed in the magnetic field is, `F=I|veclxxvecB|=IlBsintheta`. (i) F is maximum, if `sin theta=1` or `theta=90^@` i.e. `vecl` is perpendicular to `vecB`. (ii) F is minimum, if `sin theta=0` or `theta=0^@` i.e. `vecl` is parallel to `vecB`. |
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| 300. |
Explain the rule with illustration, related to the direction of current through a linear conductor and magnetic field. |
| Answer» In case of a linear conductor carrying current, the magnetic field is in the form of concentric circular magnetic lines of force whose centre lies on the conductor. These magnetic lines of force are lying in a plane perpendicular to the straight conductor. The direction of magnetic lines of force can be given by Right Hand Thumb Rule. For details, | |