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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.
| 201. |
A magnetic field exerts no force onA. a stream of electronsB. a stream of protonsC. unmagnetised piece of ironD. stationary charge |
| Answer» Correct Answer - 4 | |
| 202. |
a `2A` current is flowing through a circular coil of radius `10 cm` containing `100 ` turns. Find the magnetic flux density at the centre of the coil.A. `0.126xx10^(-2)`B. `1.26xx10^(-2)`C. `1.26xx10^(-4)`D. `1.26xx10^(-5)` |
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Answer» Correct Answer - 1 `B=(mu_(0)ni)/(2r)` |
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| 203. |
In a helium dilution refrigerator `.^(3)He and .^(4)He` are mixed in a special chamber to obtain extremely low temperature. A Bainbridge mass spectrometer is used to measure the ratio of the two isotopes (a) If the spectrometer were used with 100 V `cm^(-1)` between the plates and a magnetic field of 0.2 T, what would be the speed of an ion that can pass through the velocity filter? (b) If the velocity-filter exit slit were 1 mm wide, could this machine resolve the two isotopes ? |
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Answer» (a) `5 xx 10^(4)m//s` (b) Yes |
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| 204. |
Two coaxial plane coils, each of n turns of radius a, are separated by a distance a. calcualte the magnetic field on the axis at the point midway between them when a current l flows in the same sense through each coil. Electrons in a colou television tube are accelerated through a potential difference of 25 kV and then deflected by `45^(@)` in the magnetic field between the two coils described above. If a is 100 mm and the maximum current available for the coils is 2A, estimate the number of turns which the coils must have. |
| Answer» Approximately 200 turns | |
| 205. |
A long straight wire carries an electric current of `2A`. The magnetic induction at a perpendicular distance of `5m` from the wire is `( mu_(0)4 pi xx 10^(7)Hm^(-1))`A. `4xx10^(-8)T`B. `8xx10^(-8)T`C. `12xx10^(-8)T`D. `16xx10^(-8)T` |
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Answer» Correct Answer - 2 `B=(mu_(0)i)/(2pir)` |
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| 206. |
What is the force acting on charge (q) moving in a direction perpendicular to a magnetic field (B) with velocity v?A. `=(q)/(vec(V)xxvec(B))`B. `(vec(V)xxvec(B))/(q)`C. `q(vec(V)xxvec(B))`D. `(vec(V).vec(B))q` |
| Answer» Correct Answer - 3 | |
| 207. |
A current carrying wire produces in the neighbourhoodA. Electric and magnetic fieldsB. Electric field onlyC. Magnetic field onlyD. No field |
| Answer» Correct Answer - 3 | |
| 208. |
A doubly ionised `He^(+2)` atom travels at right angles to a magnetic field of induction `0.4T` with a velcoity of `10^(5)ms^(-1)` describing a circle of radius `r`. A proton travelling with same speed in same direction in the same field will describe a circle of radius.A. `r//4`B. `r//2`C. `r`D. `2r` |
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Answer» Correct Answer - 2 `r=(mv)/(Bq)` |
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| 209. |
A galvanometer has a resistance of `49Omega`.If `2%` of the main current is to be passed through the meter, The value of the shunt will beA. `2 Omega`B. `1 Omega`C. `(1)/(2)Omega`D. `(1)/(4)Omega` |
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Answer» Correct Answer - 2 `i_(g)=(iS)/(G+S)` |
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| 210. |
A wiere of length `L` is shaped into a cricle and then bent in such a way that the two semi`-` circle are perpendicular. The magnetic moment of the system when current `I` flows through the system isA. `(sqrt(2)iL^(2))/(8pi)`B. `(sqrt(3)iL^(2))/(4pi)`C. `(iL^(2))/(4pi)`D. `(iL^(2))/(2pi)` |
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Answer» Correct Answer - 1 `B=(mu_(0)iphi)/(2r)` |
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| 211. |
An electric current passes through a long straight wire. At a distance 5 cm from the wire the magnetic field is `B`. The field at 20 cm from the wire would beA. `2B`B. `B//4`C. `B//2`D. `B` |
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Answer» Correct Answer - 2 `Balpha(1)/(r)` |
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| 212. |
Statement-1 Helmholtz coil is used to obtain uniform magnetic field Statement-2 Magnetic field lines always form closed loop Statement-3 Magnetic field on the line of a straight current carrying wire is always zeroA. FTFB. FFTC. TTTD. TTF |
| Answer» Correct Answer - C | |
| 213. |
Figure shows sections of a long current carrying wires. At origin the current is devided in two equal parts. All section lies in xy plane. If the net magnetic field at P is `(mu_(0)lx)/(8pi) (sqrt2 + 1)^(2)`, then find out the value of x. |
| Answer» Correct Answer - 2 | |
| 214. |
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 - 4 `B=(mu_(0)i)/(2pir)` |
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| 215. |
Each of two long parallel wires carries a constant current `i` along the same direction . The wires are separated by a distance `2I`. The magnitude of resultant magnetic induction in the symmetric plane of this system located between the wires at a distance `R` from each wire will beA. `(mu_(0)i)/(piR)`B. zeroC. `(mu_(0)i)/(pisqrt(R^(2)-l^(2)))`D. `(mu_(0)i)/(piR)sqrt(1-(l^(2))/(R^(2)))` |
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Answer» Correct Answer - 2 `B_(1)=(mu_(0)i)/(2piR),B_(2)=(mu_(0)i)/(2pir)` |
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| 216. |
A straight vertical conductor carries a current. At a point `5cm` due north of it, the magnetic induction is founded to be `20muT` due east. The magnetic induction at a point `10cm` east of its will beA. `5mu T` northB. `10muT` northC. `5muT` southD. `10muT` south |
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Answer» Correct Answer - 4 `B=(mu_(0)i)/(2pir)` |
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| 217. |
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 lt1`D. `ac=b` |
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Answer» Correct Answer - 1 `R=(2rm)/(qB),R=(mv_|_)/(qB)`p itch`=Txxcos45^(@)` |
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| 218. |
In a circuit 5 percent of total current passes through a galvanometer. If resistance of the galvanometer is G then value of the shunt isA. `19G`B. `G//19`C. `20G`D. `G//20` |
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Answer» Correct Answer - 2 `S=(G)/(n-1)` |
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| 219. |
Tha scale of a galvanometer is divided into `150` equal divisions. The galvanometer be designed to read (i) `6` A per division and (ii) `1 V` per division?A. `S=8.3xx10^(-5)Omega, R=9995Omega`B. `S=8.3xx10^(-2)Omega,r=995Omega`C. `S=4.3xx10^(-5)Omega,R=995Omega`D. `R=8.3xx10^(-5)Omega,S=995Omega` |
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Answer» Correct Answer - 1 Here, `I_(g)=((1mA)/(10 d iv))(150di v)=15mA` `V_(g)=((1mA)/(2di v))(150di v)=75mV` `G=(V_(g))/(I_(g))=(75mV)/(15mA)=5 Omega` `(a) I (` current to be measured `)` `=((6A)/(di v))(150di v)=900A` As `n=(I)/(I_(g))=(900A)/(15mA)=(900A)/(15xx10^(-3)A)=6xx10^(4)` `S=(G)/((n-1))=(5Omega)/((6xx10^(4)-1))~~(5Omega)/(6xx10^(4))=8.3xx10^(-5)Omega` `(b)V (` voltage to be measured `)` `=((1V)/(di v))(150d i v)=150V` As `n=(V)/(V_(g))=(150V)/(75mV)=(150V)/(75xx10^(-3)V)=2xx10^(3)` `R=G(n-1)=5Omega(2xx10^(3)-1)` `=5 Omega(2000-1)=9995Omega` |
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| 220. |
If a voltmeter, in advertently mistaken for an ammeter, were inserted into the circuit, the currentA. increasesB. remains sameC. decreases because its reduction factor increasesD. becomes zero |
| Answer» Correct Answer - 3 | |
| 221. |
The resistance of an ideal voltmeter isA. ZeroB. infinityC. `1000Omega`D. `10000Omega` |
| Answer» Correct Answer - B | |
| 222. |
In the previous example find the mass of the wire if the magnetic field is equal to the earth magnetic field `(4 xx 10^(-5) T)`, current in the wire is 5 A and length of the wire is 1m. The wire remains suspended in mid air in equilibrium. (take `g = 10 m//s^(2)`) |
| Answer» Correct Answer - 20 mg | |
| 223. |
Assertion: Cyclotron does not accelerate. Reason: Mass of the electron is very small.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - A Cyclotron is not suitable for accelerating electrons. Due to the small mass, the speed of electrosn increaes rapdily. Likewise, due to quick relativistic variation in their mass, the electrons get out of step with the oscillating electric field. |
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| 224. |
The cyclotron is a device which is used to accelerate charged particles such as protons, deutrons, alpha particles, etc. to very high energy. The principle on which a cyclotron works is based on the fact that an electric field can accelerate a charged particle and a magnetic field can throw it into a circular orbit. A particle of charge `+q` experiences a force qE in an electric field E and this force is independent of velocity of the particle. The particle is accelerated in the direction of the magnetic field. On the other hand, a magnetic field at right angles to the direction of motion of the particle throws the particle in a circular orbit in which the particle revolves with a frequency that does not depend on its speed. A modest potential difference is used as a sources of electric field. If a charged particle is made to pass through this potential difference a number of times, it will acquire an enormous by large velocity and hence kinetic energy. Cyclotron is not suitable for acceleratingA. ElectronB. ProtonsC. DeutronsD. Alpha particles |
| Answer» Correct Answer - A | |
| 225. |
The cyclotron is a device which is used to accelerate charged particles such as protons, deutrons, alpha particles, etc. to very high energy. The principle on which a cyclotron works is based on the fact that an electric field can accelerate a charged particle and a magnetic field can throw it into a circular orbit. A particle of charge `+q` experiences a force qE in an electric field E and this force is independent of velocity of the particle. The particle is accelerated in the direction of the magnetic field. On the other hand, a magnetic field at right angles to the direction of motion of the particle throws the particle in a circular orbit in which the particle revolves with a frequency that does not depend on its speed. A modest potential difference is used as a sources of electric field. If a charged particle is made to pass through this potential difference a number of times, it will acquire an enormous by large velocity and hence kinetic energy. The working of a cyclotron is based on the fact thatA. The force experienced by a charged particles in an electric field is independent of its velocityB. The radius of the circular orbit of a charged particle in a magnetic field increase with increase in its speedC. The frequency of revolution of the particle along the circular path does not depend on its speedD. All of the above |
| Answer» Correct Answer - D | |
| 226. |
Velocity and acceleration vectors of a charged particle moving in a magnetic field at some instant are `vec(v)=3hat(i)+4hat(j)` and `vec(a)=2hat(i)+xhat(j)`. Selcet the wrong alternative.A. `x=-1.5`B. `x=3`C. magnetic field is along `z-` directionD. Kinetic energy of the particle is constant |
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Answer» Correct Answer - 2 `vec(F).vec(v)=0` |
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| 227. |
If we double the radius of a current carrying coil keeping the current unchanged, the magnetic field at its centreA. becomes four timesB. doubledC. remains unchangedD. halved |
| Answer» Correct Answer - 4 | |
| 228. |
Magnetic field at the centre of a circualr loop of area A carrying current I is B. What is the magnetic moment of this loop?A. `((BA)^(2))/(mu_(0)pi)`B. `(BAsqrtA)/(mu_(0))`C. `(BA sqrtA)/(mu_(0)pi)`D. `(2BAsqrtA)/(mu_(0)sqrtpi)` |
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Answer» Correct Answer - D Let r be the radius of the circular loop. `therefore" "A=pir^(2) or r=sqrt((A)/(pi))` Magnetic field at the centre of the loop is `B=(mu_(0)I)/(2r)=(mu_(0)I)/(2sqrt((A)/(pi)))orI=(2B)/(mu_(0))sqrt((A)/(pi))` The magnetic moment of the loop is `M=IA=(2B)/(mu_(0))sqrt((A)/(pi))A=(2BAsqrtA)/(mu_(0)sqrtpi)` |
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| 229. |
A closed current-carrying loop having a current I is having area A. Magnetic moment of this loop is defined as `vec(mu) = vec(IA)` where direction of area vector is towards the observer if current is flowing in anticlockwise direction with respect to the observer. If this loop is placed in a uniform magnetic field `vec(B)`, then torque acting on the loop is given by `vec(tau) = vec(mu) xx vec(B)`. Now answer the following questions: Consider the situation shown in Fig., ring is having a uniformly distributed positive charge. Magnetic field is perpendicular to the axis of ring. Now ring is rotated in anticlockwise direction as seen from left hand side direction of magnetic torque acting on the ring is A. Parallel toB. Parallel to axis of ringC. perpendicular to axis and magnitic fieldD. Coming out of plane of paper |
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Answer» Correct Answer - C `vec(tau)` is perpendicular `vec(M)` and `vec(B)` |
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| 230. |
A closed current-carrying loop having a current I is having area A. Magnetic moment of this loop is defined as `vec(mu) = vec(IA)` where direction of area vector is towards the observer if current is flowing in anticlockwise direction with respect to the observer. If this loop is placed in a uniform magnetic field `vec(B)`, then torque acting on the loop is given by `vec(tau) = vec(mu) xx vec(B)`. Now answer the following questions: A uniformly charged insulating ring is rotated in a uniform magnetic field about its own axis, thenA. Ring will experience a magnetic forceB. Ring must be experience a magnetic torqueC. Ring may experience a magnetic torqueD. None of the above |
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Answer» Correct Answer - C `tau=MB sin theta`, Rotating ring will behave as a magnetic pole. |
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| 231. |
A closed current-carrying loop having a current I is having area A. Magnetic moment of this loop is defined as `vec(mu) = vec(IA)` where direction of area vector is towards the observer if current is flowing in anticlockwise direction with respect to the observer. If this loop is placed in a uniform magnetic field `vec(B)`, then torque acting on the loop is given by `vec(tau) = vec(mu) xx vec(B)`. Now answer the following questions: Let ring in the above question is having a radius R and a charge Q is uniformly distributed over it. Ring is rotated with a constant angular velocity `(omega)` as mentioned above. Torque acting on the ring due to magnetic force isA. `(QR^(2)omegaB)/(2)`B. `(piR^(2)qB)/(2omega)`C. `(qomegaR^(2)B)/(2pi)`D. None of the above |
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Answer» Correct Answer - A Torque `=((q omegaR^(2))/(2))B(tau=MB)` |
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| 232. |
An 8 cm long wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. If the magnetic field inside the solenoid is 0.3 T, then magnetic force on the wire isA. 0.14NB. 0.24NC. 0.34ND. 0.44N |
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Answer» Correct Answer - B `F=Il B sin theta` Here, `theta=90^(@),I=10A,` `l= 8cm = 8xx10^(-2)m, B=0.3T` `therefore" "F=10xx8xx10^(-2)xx0.3xxsin 90^(@)=0.24N` |
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| 233. |
An electron is projected with uniform velocity along the axis of a current carrying long solenoid. Which of the following is true?A. The electron will be accelerated along the axis.B. The electron path will be circular about the axis.C. The electron will experience a force at `[email protected]` to the axis and hence execute a helical path.D. The electron will contnue to move with uniform velocity along the axis of the solenoid. |
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Answer» Correct Answer - D Let the electron ( e) is projected with a uniform velocity ( v) in a uniform magnetic field B. The magnetic velocity (v ) in a uniform magnetic field B. The magnitude of force on it is `|vecF|=-e|vecv xx vecB| =-evB sintheta` As `theta = 0^(@), |vecF| =-evBsin 0^(@)=0` Hence the electron will continue to move with a uniform velocity along the axis of the solenoid. |
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| 234. |
In a current carrying long solenoid, the field produced does not depend uponA. Number of turns per unit lengthB. Current flowingC. Radius of the solenoidD. All of these |
| Answer» Correct Answer - C | |
| 235. |
If a long hollow copper pipe carriers a direct current, the magnetic field associated with the current will be:A. inside the pipe onlyB. Outside the pipe onlyC. Neither inside nor outside the pipe.D. Both inside and outside the pipe. |
| Answer» Correct Answer - 2 | |
| 236. |
If a long hollow copper pipe carriers a direct current, the magnetic field associated with the current will be:A. Inside the pipe onlyB. Outside the pipe onlyC. Both inside and outside the pipeD. Neither inside nor outside the pipe |
| Answer» Correct Answer - B | |
| 237. |
A particle carrying a charge equal to `100` times the charge on an electron is rotating per second in a circular path of radius `0.8 metre`. The value of the magnetic field produced at the centre will be (`mu_(0)=` permeability for vacuum)A. `10^(-7)//mu_(0)`B. `10^(-17)mu_(0)`C. `10^(-6)mu_(0)`D. `10^(-15)//mu_(0)` |
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Answer» Correct Answer - 2 `B=(mu_(0)n e)/(2r)` where `n` is the frequency |
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| 238. |
Magnetic field induction at the center of a circular coil of radius `5cm` and carrying a current `0.9A` is `(` in `S.I.` units in`)(in_(0)=` absolute permitivity of air in `S.I.` units `:` velocity of light `=3xx10^(8)ms^(-1))`A. `(1)/(in_(0)10^(16))`B. `(10^(16))/(in_(0))`C. `(in_(0))/(10^(16))`D. `10^(16)in_(0)` |
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Answer» Correct Answer - A `B=(mu_(0)ni)/(2r),c=(1)/(sqrt(mu_(0)E_(0)))` |
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| 239. |
The magnetic induction at a point at a large distance `d` on the axial line of circular coil of small radius carrying current is `120muT` . At a distance `2d` the magnetic induction would beA. `60muT`B. `30 mu T`C. `15mu T`D. `240 mu T` |
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Answer» Correct Answer - C `Balpha(1)/(x^(3))` |
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| 240. |
In an inertial frame of reference, the magnetic force on a moving charged particle is F. Its value in another inertial frame of reference will beA. remained sameB. changed due to change in the amount of chargeC. changed due to change in velocity of charged particleD. changed due to change in field direction |
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Answer» Correct Answer - C `vecF=q(vecv xx vecB) therefore F="qv B sin "theta` which shows magnetic field is velocity dependent due to which it differs from one inertial frame to another. |
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| 241. |
A horizontal wire carries `200amp` current below which another wire of linear density `20xx10^(-5)kgm^(-1)` carrying a current is kept at `2 cm` distance. If the wire kept below hangs in air. The current in this wire isA. `100A`B. `9.8A`C. `98A`D. `48A` |
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Answer» Correct Answer - C `(mu_(0)i_(1)i_(2))/(2pir)=(mg)/(l)` |
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| 242. |
A square loop of side `L` carries a current `I`. Another smaller square loop of side `l(l lt lt L)` carrying a current `i` is placed inside the bigger loop such that they are coplanar with their centre coinciding. If the currents in the loops are in the same direction, the magnitude of the torque on the smaller loop isA. `(mu_(0)Iil^(2))/(sqrt(2)piL)`B. `(mu_(0)Iil^(2))/(2piL)`C. `(mu_(0)Iil^(2))/(sqrt(3)2piL)`D. Zero |
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Answer» Correct Answer - 4 `tau=BiAncos90^(@)` |
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| 243. |
Five long wires A, B, C, D and E, each carrying current I are arranged to form edges of a pentagonal prism as shown in figure. Each carries current out of the plane of paper. (a) What will be magnetic induction at a point on the axis O? Axis is at a distance R from each wire. (b) What will be the field if current in one of the wires (say A) is switched off? (c) What if current in one of the wire (say) A is reversed?A. equal to zeroB. less than zeroC. more than zeroD. infinite |
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Answer» Correct Answer - A The resultant magnetic field induction at O due to current through all the five wires will be zero. |
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| 244. |
A long straight wire carrying current of 30 A rests on a table. Another wire AB of length 1 m, mass 3 g carries the same current but in the opposite direction, the wire AB is free to slide up and down. The height upto which AB will rise isA. 0.6 cmB. 0.7 cmC. 0.4 cmD. 0.5 cm |
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Answer» Correct Answer - A Here `I_(1)=I_(2) = 30A, l = 1m, m = 3 g = 3xx10^(-3)kg` In equilibrium position , `mg=(mu_(0))/(4pi)(2I_(1)I_(2)l)/(h), h=(mu_(0))/(4pi)xx(2I_(1)I_(2)l)/(mg)` `h=10^(-7)xx(2xx30xx30xx1)/(3xx10^(-3)xx10)=0.6cm` |
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| 245. |
The magnetic moment of a current (i) carrying circular coil of radius (r) and number of turns (n) varies asA. `1//r^(2)`B. `1//r`C. rD. `r^(2)` |
| Answer» Correct Answer - D | |
| 246. |
Electrical resistance of certain materials, known as superconductors, changes abruptly from a nonzero value of zero as their temperature is lowered below a critical temperature `T_(C) (0)`. An interesting property of super conductors is that their critical temperature becomes smaller than `T_(C) (0)` if they are placed in a magnetic field, i.e., the critical temperature `T_(C) (B)` is a function of the magnetic field strength B. The dependence of `T_(C) (B)` on B is shown in the figure. . In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields `B_1`(solid line) and `B_2` (dashed line). If `B_2` is larget than `B_1` which of the following graphs shows the correct variation of R with T in these fields? |
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Answer» Correct Answer - A From give graph between `T_(c) (B) vs B`, it is clear that as magnetic field increases critical temperature decreases. For `Q.1` if `B_(2)gtB_(1)` so `T_(c)(B_(2))ltT_(c)(B_(1))` and `T_(c)` is the temperature at which resistance of superconductor becomes zero . The inequality in option `(a)`. |
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| 247. |
The magnetic moment of a current I carrying circular coil of radius r and number of turns N varies asA. `1/(r^(2))`B. `1/r`C. rD. `r^(2)` |
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Answer» Correct Answer - D Magnetic moment, M = NIA = `NI pi r^(2)` i.e. `M prop r^(2)` |
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| 248. |
A circular coil of radius 10 cm having 100 turns carries a current of 3.2 A. The magnetic field at the center of the coil isA. `2.01 xx 10^(-3)T`B. `5.64 xx 10^(-3)T`C. `2.64 xx 10^(-3)T`D. `5.64 xx 10^(-3)T` |
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Answer» Correct Answer - A As `B=(mu_(0)NI)/(2R)," Here N "= 100, I = 3.2 A,` `R = 10 cm = 10xx10^(-2)m` `therefore" "b=(4pixx10^(7)xx100xx3.2)/(2xx0.1)=2.01xx10^(-3)T` |
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| 249. |
A circular coil of radius `5 cm ` has 169 turns carries a current of `2.6 A`. The magnetic induction at a point on the axis at a distance of `12 cm` from the centre of the coil isA. `1T`B. `4.2T`C. `3.14xx10^(-4)`D. `3.14xx10^(-2)T` |
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Answer» Correct Answer - C `B=(mu_(0)nir^(2))/(2(r^(2)+x^(2))^(3//2))` |
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| 250. |
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 magnitude of the force required to move the conducting rod at constant speed 5cm/s at the same instant `t=2s`, is equal toA. `0.16N`B. `0.12N`C. `0.08N`D. `0.06N` |
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Answer» Correct Answer - C `(C )` At `t=2s`, length of the wire `=(2xx30cm)+20cm =0.8m` Resistance of the wire `=0.8 Omega` Current through the rod `=(0.08)/(0.8)=(1)/(10)A` Force on the wire `=i//B` `=(1)/(10)xx(0.2)xx4=0.08N` Same force is applied on the rod in opposite direction to make net force zero. |
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