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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. |
Figure shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance `1Omega//m`. Position of the conducting rod at `t=0` is shown. A time dependent magnetic field `B=2t` tesla is switched on at `t=0` The current in the loop at `t=0` due to induced emf isA. `0.16A,`clockwiseB. `0.08A,` clockwiseC. `0.08A,` anticlockwiseD. zero |
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Answer» Correct Answer - A `(A)(dB)/(dt)=2T//s` `E=-(AdB)/(dt)=-800xx10^(-4)ms^(2)xx2=-0.16V` `i=(0.16)/(1Omega)=0.16A,` clockwise |
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| 252. |
A 200 turn closely wound circular coil of radius 15 cm carries a current of 4 A. The magnetic moment of this coil isA. `36.5A m^(2)`B. `56.5A m^(2)`C. `66.5A m^(2)`D. `108A m^(2)` |
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Answer» Correct Answer - B The magnetic moment is given by `|vecm|=NIA=NI pi r^(2)=200xx4xx3.14xx(15xx10^(-2))^(2)` `=200xx4xx3.14xx15xx15xx10^(-4)="56.5 A m"^(2)` |
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| 253. |
The magnetic moment associated with a circular coil of 35 turns and radius 25 cm, if it carries a current ofA. `72.2A m^(2)`B. `70.5A m^(2)`C. `74.56A m^(2)`D. `75.56A m^(2)` |
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Answer» Correct Answer - D Given N = 35, r = 25 cm `=25xx10^(-2)m,` I = 11 A Then magnetic moment associated with this circular coil `M = NIA = NIpi r^(2)=35xx11xx3.14xx(25xx10^(-2))^(2)` `=75.56"A m"^(2)` |
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| 254. |
A uniform conducting wire of length length 10a and resistance R is wound up into four turn as a current carrying coil in the shape of equilateral triangle of side a. If current I is flowing 4 through the coil then the magnetic moment of the coil is A. `(sqrt3)/(2)a^(2)I`B. `(a^(2)I)/sqrt3`C. `sqrt3a^(2)I`D. `(2a^(2)I)/(sqrt3)` |
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Answer» Correct Answer - C Magnetic moment M = NIA `=4xxIxxa^(2)(sqrt3)/(4)=sqrt3a^(2)I` |
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| 255. |
A current I flows in a rectangularly shaped wire whose centre lies at `(x_0,0,0)` and whose vertices are located at the points A `(x_0+d,-a,-b),B(x_0-d,a,-b),C(x_0-d,a,+b) and D(x_0+d,-a,b)` respectively. Assume that a,b,dltlt`x_0`. Find the magnitude of magnetic dipole moment vector of the rectangular wire frame in `J//T`. (Given, b=10 m,I=0.01A, d=4m,a=3m).A. `2 J T^(-1)`B. `4 J T^(-1)`C. `3 J T^(-1)`D. `9 J T^(-1)` |
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Answer» Correct Answer - A Magnetic moment of a current carrying loop, `vecmu = IvecS` Area of the loop, `vecS = vec(AB) xx vec(BC)` Here, `vec(AB) = -2d hati +2ahatj , vec(BC) = 2b hatk` `therefore vecS=(-2d hati + 2a hatj) xx (2b hatk) = 4bd hati + 4ab hati` `therefore |vecmu|= I|vecS|=4Ib sqrt(a^(2)+d^(2))` `=4 xx 0.01 xx 10 xx sqrt(3^(2) + 4^(2)) = 0.4 xx 5 = 2J T^(-1)` |
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| 256. |
On an inclined plane, two conducting rails are fixed along the line of greatest slope. A conducting rod AB is placed on the radils so that it is horizontal. A vertically upward uniform magnetic field is present in the region. A current i flows through AB from B to A. The mass of AB is m and the coefficient of friction between the rod and rails is `mu`. Find the minimum and maximum value of B, so that the rod can stay in equilibrium. Assume that `mu lt tan theta` |
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Answer» `B_("min") = (mg (sin theta - mu cos theta))/(lL (cos theta + mu sin theta))` `B_("max") = (mg (sin theta + mu cos theta))/(IL (cos theta - mu sin theta))` For equilibrium `B_("min") le B le B_("max")` |
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| 257. |
An ammeter and a voltmeter of resistance `R` connected in seires to an electric cell of negligible internal resistance. Their readings are `A` and `V` respecitvely. If another resistance `R` is connected in parallel with the voltmeterA. Both `A` and `V` increasesB. Both `A` and `V` decreasesC. `A` decreases but `V` increasesD. `A` increases but `V` decreases |
| Answer» Correct Answer - 4 | |
| 258. |
Statement-1 Magnetic force is always onon-central while the electric force may be central Statement-2 magnetic field can acclerate a charged particle Statement-3 Consider a long, straight wire of radius R, carrying a current distributed uniformly over its cross-section. The magnitude of the magnetic field is maximum at surface of the wire.A. FFFB. FTFC. TFTD. TTT |
| Answer» Correct Answer - D | |
| 259. |
Statement-1 Magnetic field inside an ideal solonoid is uniform Statement-2 Magnetic field outside an ideal solonoid is zero Statement-3 Magnetic field at centre of an ideal solonoid is twice the magnetic field at the endsA. TTTB. TTFC. FFFD. FFT |
| Answer» Correct Answer - A | |
| 260. |
Statement-1 If a positive charge is thrown parallel to a current carrying wire it will be attracted by the wire Statement-2 If a negative charge is thrown antiparallel to a current carrying wire it will be repelled by the wire Statement-3 A current carrying wire can apply to a force on a charge placed near itA. FTTB. FTFC. TFFD. TTF |
| Answer» Correct Answer - C | |
| 261. |
A closed circuit is in the form of a regular hexagon of side r. If the circuit carries current I, what is the magnetic field induction at the centre of the hexagon?A. `(sqrt(3)mu_(0)I)/(4pia)`B. `(sqrt(3)mu_(0)I)/(2pia)`C. `(sqrt(3)mu_(0)I)/(3pia)`D. `(sqrt(3)mu_(0)I)/(pia)` |
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Answer» Correct Answer - 4 Magnetic field at `O` due to `AB, i.e` `B=(mu_(0)I)/(4pid)(sinphi_(1)+sinphi_(2))` |
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| 262. |
A ring carrying current l lies in x-z plane as shown. A short magnetic dipole with dipole moment directed along x-axis is fixed at origin. If the ring is free to move anywhere, then it will A. Rotate clockwise about z-axisB. Rotate anticlockwise about z-axisC. Move along z-axisD. Move along y-axis |
| Answer» Correct Answer - A | |
| 263. |
If a long straight wire carries a current of 40 A, then the magnitude of the field B at a point 15 cm away from the wire isA. `5.34xx10^(-5)T`B. `8.34xx10^(-5)T`C. `9.6xx10^(-5)T`D. `10.2xx10^(-5)T` |
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Answer» Correct Answer - A I = 40 A r = 15 cm `=15xx10^(-2)m` `therefore" "B=(mu_(0)I_("enc"))/(2piR)=2xx10^(-7)xx(75)/(3)=5xx10^(-6)T` |
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| 264. |
A long straight wire in the horizontal plane carries as curret of 75 A in north to south direction, magnitude and direction of field B at a point 3 m east of the wire isA. `4xx10^(-6)T`, vertical upB. `5xx10^(-6)T`, vertical downC. `5xx10^(-6)T`, vertical upD. `4xx10^(-6)T`, vertical down |
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Answer» Correct Answer - C From Ampere circuital law `ointvecB.vec(dl)=mu_(0)I_("enc")` `Bxx2piR=mu_(0)I_("enc")` `B=(mu_(0)I_("enc"))/(2piR)=2xx10^(-7)xx(75)/(3)=5xx10^(-6)T` The direction of field at the given point will be vertical up determined by the screw rule or right hand rule. |
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| 265. |
Three very long straight conductors are arranged parallel to each other in a plane and have resistances in the ration `1:2:3` . They are connected in parallel to a battery of negligible internal resistance such that the currents in all three wires are in the same direction. The distance between the first two conductors is `x` and the distance between the second and third conductors is `y`. If the middle conductors is in equilibrium, the ratio `x:y` isA. `1:3`B. `3:1`C. `1:sqrt(3)`D. `sqrt(3):1` |
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Answer» Correct Answer - 2 `F=(mu_(0)i_(1)i_(2)L)/(2pir)` |
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| 266. |
Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field `vec(B)=B_(0)hat(K)`A. They have equal `z-`components of momenta.B. They must have equal charges.C. They necessarily represent a particle-antiparticle pair.D. The charge to mass ratio satisfy: `((e)/(m))_(1)+((e)/(m))_(2)=0`. |
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Answer» Correct Answer - D In the present situation the charge to mass ratio (e/m) of these two particles is same and charges on them are of opposite character. Hence, the situation given in option (d) holds good. |
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| 267. |
Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field `vec(B)=B_(0)hat(K)`A. They have equal `z-` components of momentaB. They must have equal chargesC. They necessarily represent a partical `-` antipartical paritD. The charge to mass ratio satisfy `:` `(e//m)_(1)+(e//m)_(2)=0` |
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Answer» Correct Answer - 4 Since the charged particles traverse identical halical paths in a completely opposite sense in a uniform magentic field `B`, their radii` (r=mv sin theta//eB)` and frequencies `(v=eB//2pim)` must be equal For this `(e//m)_(1)=-(e//m)_(2) or (e//m)_(1)+(e//m)_(2)=0` |
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| 268. |
The gyromagnetic ratio of an electron in sodium atom isA. depending upon the atomic number of the atomB. depending upon the shell number of the atomC. independent of that orbit it is inD. having positive value |
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Answer» Correct Answer - C `"Gyromagnetic ratio "=("magnetic moment")/("angular momentum")` `=("neg / "4pim)/(nh//2pi)=(e)/(2m)="constant"` Hence, gyromagnetic ratio is not dependent on the orbit and for electron,it is having negative value due to negative charge on electron. |
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| 269. |
Acceleration experienced by a particle with specific charge` 1xx10^(7) C//kg` when fired perpendicular to a magnetic field of induction `100 mu T` with a velocity `10^(5)ms^(-1)` isA. `10^(8)ms^(-2)`B. `10^(-6)ms^(-2)`C. `10^(14)ms^(-2)`D. `10^(-8)ms^(-2)` |
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Answer» Correct Answer - 1 `F=qvB` |
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| 270. |
When two electrons enter into a magnetic field with different velocieis, they deflect in different circular paths, in such a way that the radius of one path is double that of the other. `1X10^(7)ms^(-1)` is the velocity of the electron in smaller of radius `2xx10^(3)m` . The velocity of electron in the other circular path is `:`A. `4xx10^(7)ms^(-1)`B. `4xx10^(6)ms^(-1)`C. `2xx10^(7)ms^(-1)`D. `2xx10^(6)ms^(-1)` |
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Answer» Correct Answer - 3 `v=(qBr)/(m)` |
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| 271. |
Two particles having same charge and `KE` enter at right angles into the same magnetic field and travel in circular paths of radii `2 cm` and `3cm` respectively. The ratio of their masses is .A. `2:3`B. `3:2`C. `4:9`D. `9:4` |
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Answer» Correct Answer - 3 `r=(mv)/(Bq)=(sqrt(2mk))/(Bq)` |
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| 272. |
Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii `R_1 and R_2,` respectively. The ratio of masses of X and Y isA. `(R_(1)//R_(2))^(1//2)`B. `(R_(2)//R_(1))`C. `(R_(1)//R_(2))^(2)`D. `(R_(1)//R_(2))` |
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Answer» Correct Answer - 3 `r=(mV)/(Bq)` |
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| 273. |
Two charges `q_(1) and q_(2)` having same magneitude of charge are moving parallel to each other and they enter into a region of uniform magnetic field as shown. If they have same mass and the time spent by them in the magnetic field are `t_(1) and t_(2)` respectively, then A. For `q_(1) = +- q_(2), t_(1) = t_(2)`B. For `q_(1) = q_(2), t_(1) = t_(2)`C. For `q_(1) gt 0, q_(2) lt 0, t_(1) lt t_(2)`D. For `q_(1) lt ), q_(2) gt 0, t_(1) lt t_(2)` |
| Answer» Correct Answer - B::D | |
| 274. |
A straight wire carrying a current of 13 A is bent into a semi-circular arc of radius 2 cm as shown in figure. The magnetic field is `1.5 xx 10^(-4)T` at the centre of arc, then the magnetic field due to straight segment is A. `1.5 xx 10^(-4)T`B. `2.5xx10^(-4)T`C. zeroD. `3xx 10^(-4)T` |
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Answer» Correct Answer - C Since dl and r for each element of the straight segments are parallel. Therefore `vec(dl)xx vecr=0` Hence, B is also zero. |
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| 275. |
If a shunt is to be applied to a galvanometer of resistance `50Omega` so that only `5%` of total current passes through the galvanometer. The resistance of shunt should beA. `1.63Omega`B. `4.2Omega`C. `3.5Omega`D. `2.63Omega` |
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Answer» Correct Answer - 4 `S=(G)/(n-1)` |
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| 276. |
The value of current in the given circuit if the ammeter is a galvanometer with a resistance `R_(G)=50Omega` is A. 0.048 AB. 0.023 AC. 0.061 AD. 0.094 A |
| Answer» Correct Answer - D | |
| 277. |
In a cyclotron, a charged particleA. undergoes acceleration all the time.B. speeds up between the dees because of the magnetic field.C. speeds up in a dee.D. slows down within a dee and speeds up between dees. |
| Answer» Correct Answer - A | |
| 278. |
In a cyclotron , if the frequency of proton is `5MHz`, the magnetic field necessary for resosnance isA. `0.528T`B. `2.32T`C. `0.389T`D. `0.327T` |
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Answer» Correct Answer - D `f=(Bq)/(2pim)` |
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| 279. |
Cyclotron is adjusted to give proton beam, magnetic induction is `0.15 wbsm^(-2)` and the extreme radius is `1.5m` The energy of emergent proton ini `MeV` will beA. `34.2`B. `3.42`C. `2.43`D. `24.3` |
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Answer» Correct Answer - 3 `r=(mv)/(Bq),KE=(1)/(2)mv^(2)` |
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| 280. |
If in circular coil of radius `R`, current `I` is flowing and in another coil `B` of radius `2R` a current `2I` is flowing , then the raatio of the magnetic fields `B_(A) and B_(B)`, produced by them will beA. `1`B. `2`C. `(1)/(2)`D. `4)` |
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Answer» Correct Answer - A KEY CONCEPT : we know that the magnetic field produced by a current carrying circular coil of radius `r` at its centre is `B = (mu_(0))/(4pi) (I)/( r) xx2 pi` Here `B_(A) = (mu_(0))/(4pi) (I)/( r) xx2 pi and B_(B) = (mu_(0))/(4pi) (2I)/( 2r) xx2 pi` rArr `(B_(A))/(B_(B)) = 1` |
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| 281. |
A cyclotron is operated at an oscillator frequency of 12 MHz and has a dee radius R 50 cm. What is the magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron?A. 0.72 TB. 0.65 TC. 0.39 TD. 0.12 T |
| Answer» Correct Answer - A | |
| 282. |
A cubical region of space is filled with some uniform electric and magnetic field. An electron enters the cube across one of its faces with velocity v and a positron enters via opposite face with velocity- v. At this instant, which one of the following is not correct?A. The electric forces on both the particles cause identical acceleration.B. The magnetic forces on both the particles cause equal acceleration.C. Both particles gain or loose energy at the same rate.D. The motion of the centre of mass is determined by B alone. |
| Answer» Correct Answer - A | |
| 283. |
A charged particle would continue to move with a constant velocity in a region wherein, which of the following conditions is not correct?A. `E=0, B ne 0`B. `E ne 0, B ne 0`C. `E ne 0, B =0`D. `E=0, B=0` |
| Answer» Correct Answer - C | |
| 284. |
A charged particle is at rest in the region where magnetic field and electric field are parallel. The particle will move in aA. Stright lineB. CircleC. EllipseD. None of these |
| Answer» Correct Answer - A | |
| 285. |
A proton (or charged particle) moving with velocity `v` is acted upon by electric field `E` and magnetic field `B`. The proton will move indeflected ifA. `E` is perpendicular to `B` and `E` parallel to `V`B. `E` is parallel to `V` and perpendicular to `B`C. `E` and `B` both are perpendicular to `V`D. `E,V` and `B` are mutually perpendicular and `V=E//B` |
| Answer» Correct Answer - 4 | |
| 286. |
A proton and an `alpha`-particle, accelerated through same potential difference, enter a region of uniform magnetic field with their velocities perpendicular to the field. Compare the radii of circular paths followed by them. |
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Answer» Let mass of proton = m, charge of proton = e Now, Mass of `alpha`-particle = 4m Chanrge of `alpha`-particle = 2e When a charge q is accelerated by V volts, it acquires a kinetic energy `E_(k) = qV` `:.` Momentum is given by `mv = sqrt(2m E_(k)) = sqrt(2mqV)` Radius `r = (mv)/(qB) or r = (sqrt(2mqV))/(qB) = sqrt((2mV)/(qB^(2))` Thus `(r_(p))/(r_(alpha)) = sqrrt((2mV)/(eB^(2))) xx sqrt((2eB^(2))/(2(4m) V)) = (1)/(sqrt2)` |
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| 287. |
An alpha particle and proton enter a region of magnetic field which is perpendicular to their direction of motion. Find ratio of radius of circle described by them if they (i) have same velocity (ii) have same momentum (iii) have same energy (iv) are accelerated by same potential difference |
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Answer» (i) `2 : 1` (ii) `1 : 2` (iii) `1 : 1` (iv) `sqrt2 : 1` |
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| 288. |
An electron and a proton enter a magnetic field at right angles to the field with the same kinetic energyA. trajectory of electron is less curvedB. trajectroy of proton is less curvedC. both are equally curvedD. both move along straight line paths |
| Answer» Correct Answer - 2 | |
| 289. |
A proton an an `alpha-`particle, moving with the same velocity, enter a uniform magnetic field, acting normal to the plane of their motion. The ratio of the radii of the circular paths descirbed by the proton and `alpha`-particle isA. `1 : 2`B. `1 : 4`C. `1 : 16`D. `4 : 1` |
| Answer» Correct Answer - A | |
| 290. |
Doubly ionized oxygen atoms `(O^(2-))` and singly-ionized lithium atoms `(Li^(+))` are travelling with the same speed, perpendicular to a uniform magnetic field. The relative atomic masses of oxygen and lithium are 16 and 7 respectively. The ratio `("radius of " O^(2-) "orbit")/("radius of "Li^(+) " orbit")` isA. `16 : 7`B. `8 : 7`C. `7 : 8`D. `7 : 16` |
| Answer» Correct Answer - B | |
| 291. |
A beam of protons with a velocity of `4 X 10 ^5 ms^(-1)` enters a uniform magnetic field of 0.3 T. The velocity makes an angle of `60^@` with the magnetic field. Find the radius of the helicla path taken by the proton beam and the pitch of the helix.A. `4.7m`B. `0.47m`C. `0.047m`D. `0.0047m` |
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Answer» Correct Answer - 3 `p=v cos theta xxT,T=(2pir)/(v sin theta)` |
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| 292. |
A beam of protons with a velocity of `4 X 10 ^5 ms^(-1)` enters a uniform magnetic field of 0.3 T. The velocity makes an angle of `60^@` with the magnetic field. Find the radius of the helicla path taken by the proton beam and the pitch of the helix. |
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Answer» Velocity component along the field `V_(||) = 4 xx 10^(5) xx cos 60^(@)` `= 2 xx 10^(5) m//s` and `v_(bot) = (4 xx 10^(5)) sin 60^(@) = 2 sqrt3 xx 10^(5) m//s` Proton will describe a circle in plane perpendicular to magnetic field with radius `r = (mv_(bot))/(qB) = ((1.67 xx 10^(-27) kg) xx (2 sqrt3 xx 10^(5) m//s))/((1.6 xx 10^(-19) C) xx (0.3 T))` `= 1.2 cm` Time taken to complete one revolution is `T = (2pi r)/(v_(bot)) = (2 xx 3.14 xx 0.012)/(2 sqrt3 xx 10^(5) m//s)` Because of `v_(||)` protons will also move in the direction of magnetic field. Pitch of helix `= v_(||) xx T` `= (2 xx 10^(5) xx 2 xx 3.14 xx 0.012)/(2 sqrt3 xx 10^(5) m//s)` `= 0.044 = 4.4 cm` |
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| 293. |
An electron having momentum `2.4 xx 10^(-23)" kg m s"^(-1)` enters a region of uniform magnetic field of 0.15 T. The field vector makes an angle of `30^(@)` with the initial velocity vector of the electron. The radius of the helical path of the electron in the field shall beA. 2mmB. 1mmC. `sqrt3/2` mmD. 0.5mm |
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Answer» Correct Answer - D The radius of the helical path of the electron in the uniform magnetic field is `r=(mv_(_|_))/(eB)=("mv sin "theta)/(eB)=((2.4xx10^(-23)"kg m s"^(-1))xxsin 30^(@))/((1.6xx10^(-19)C)xx(0.15T))` `=5xx10^(-4)m=0.5xx10^(-3)m=0.5mm` |
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| 294. |
Two particles `X` and `Y` with equal charges, after being accelerated throuhg the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii `R_(1)` and `R_(2)` respectively. The ratio of the mass of `X` to that of `Y` isA. `R_(1) //R_(2)`B. `(R_(1)//R_(2))^(2)`C. `(R_(2)//R_(1))`D. `(R_(2)//R_(2))^(2)` |
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Answer» Correct Answer - B `qV=(1)/(2)Mv^(2) or v=sqrt((2qV)/(M))` `"And "Bqv=(Mv^(2))/(R) or Bq=(Mv)/(R)=(M)/(R)sqrt((2qV)/(M))=(sqrt(2qVM))/(R)" ...(Using (i))"` `or M=(B^(2)q^(2)R^(2))/(2qV)=(B^(2)qR^(2))/(2V)` `therefore" "M prop r^(2)" "({:(because "B, q and V are same for the "),("given two particles"):})` `"Hence "(M_(1))/(M_(2))=((R_(1))/(R_(2)))^(2)` |
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| 295. |
Two long thin, parallel conductors carrying equal current in the same direction are fixed parallel to the x-axis one passing through magnetic field due to the two conductors at any point is `B` Which of the following are correct ?(A) B=0 for all points on the x-axis (B) At all points on the y-axis, excluding the origin, B has only a zcomponent. (C) At all points on the z-axis, excluding the origin, B has only an x-component. A. `B=0`, for all points on the `x-` axisB. At all points on the `y-` axis, excluding the origin, `B` has only a `z-`component.C. At all points on the `z-` axis , excluding the origin, `B` has only a `y-` component.D. `B` cannot have an `x-` component |
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Answer» Correct Answer - A,B,C,D Magnitude of magnetic field on `x-` axis form both the wire are same and opposite direction. So, B on axis is zero. |
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| 296. |
Two circular coil of radii 5cm and 10cm carry equal currents of 2A. The coils have 50 and 100 turns, respectively, and are placed in such a way that their planes as well as their centers coincide. Magnitude of magnetic field at the common centre of coils isA. `8pixx10^(-4)T` if current in the coil are in same sense.B. `4pixx10^(-4)T` if current in the coil are in opposite sense.C. Zero if currents in the coils are in opposite sense.D. `8pixx10^(-4)T` if current in the coil are in opposite sense. |
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Answer» Correct Answer - A,C `B_(1)=(mu_(0)N_(1)I)/(2r_(1))+(mu_(0)N_(2)I)/(2r_(2))` `=8000pixx10^(-7)=8pixx10^(-4)T` |
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| 297. |
The coils A and B having the radii in the ratio `1:2` carrying currents in the ratio `5:1` and have the number of turns in the ration `1:5` . The ratio of magnetic inductions at their centres is `A. `1:2`B. `2:1`C. `1:5`D. `5:1` |
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Answer» Correct Answer - 2 `B prop (n i)/( R)` |
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| 298. |
Two concentric circular coils `A` and `B` have radii `25cm` and `15cm` and carry currents `10A` and `15A` respectively. A has 24 turns and `B` has 18 turns. The direction of currents are in opposite order. The magnetic induction at the common centre of the coil isA. `120 mu_(0)T`B. `480 mu_(0)T`C. `420mu_(0)T`D. `mu_(0)` |
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Answer» Correct Answer - 3 `B=B_(1)~B_(2)` |
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| 299. |
A thin wire of length `L` is made of an insulating material. The wire is bent to form a circular loop, and a positive charge `q` is distributed uniformly around the circumference of the loop. The loop is then set into rotation with angular speed `omega` around an axis through its centre. If the loop is in the region where there is a uniform magnetic field `vec(B)` directed parallel to the plane of the loop, calculate the magnitude of the magnetic torque on the loop.A. `(q omegaL^(2)B)/(8pi^(2))`B. `(q omegaL^(2)B)/(4pi^(2))`C. `(q omegaL^(2)B)/(2pi^(2))`D. `(q omegaL^(2)B)/(pi^(2))` |
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Answer» Correct Answer - 1 `tau=mBsintheta=IABsintheta,IAB,(as theta=90^(@))` Further, as ` I=(q)/(T)=(q)/(2pi//omega)=(qomega)/(2pi),` `A=pir^(2)=pi((L)/(2pi))^(2)=(L^(2))/(4pi)` `tau=((qomega)/(2pi))((L^(2))/(4pi))B=(qomegaL^(2)B)/(8pi^(2))` |
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| 300. |
A microameter has a resistance of `100 omega and a full scale range of ` 50 muA`. It can be used as a voltmeter or as a higher range ammeter provides a resistance is added to it . Pick the correct range and resistance combination(s)A. ` 50 V ` range with `10 komega` resistance in seriesB. ` 10 V ` range with `200 komega` resistance in seriesC. ` 5 mA ` range with `1 omega` resistance in parallelD. ` 10 mA ` range with `1 omega` resistance in parallel |
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Answer» Correct Answer - B::C For `V = I_(g)(G + R) = 5xx10^(-5)[ 100 + 200, 000] = 10v` For ` I = I_(g) ((G)/(S) + 1) = 5 xx 10^(-5)[ (100)/(1) + 1] = 5 mA.` |
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