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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.
| 1. |
The variation of magnetic susceptibility `(chi)` with absolute temperature `T` for a ferromagnetic material is |
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Answer» Correct Answer - A Susceptibility of a ferromagnetic substance falls with rise of temperature `(chi=c/(T-T_(c)))` and the substance becomes paramagnetic above Curie temperature, so magnetic susceptibility becomes very small above Curie temperature. |
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| 2. |
Point out the best representation of relation between magnetic susceptibility `(Chi)` and temperature (T) for a paramagnetic materialA. B. C. D. |
| Answer» Correct Answer - D | |
| 3. |
The hysteresis loss for a specimen of iron weighing 15kg is equivalent at `300Jm^(-3)cycle^(-1)` Find the loss of energy per hour at 25 cycle `s^(-1)`. Density of iron is `7500kg m^(-3)` |
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Answer» Let Q be th energy dissipated per unit volume per hystersis cycle in the given sample. Then the total energy lost by the volume V of the sample in time t will be `" "W=QxxVxxvxxt` where v is the number of hysteresis cycle per second. Here `Q=300Jm^(-1)"cycle"^(-1),v=25"cycle"s^(-1),t=1h=3600s` Volume, `V=("Mass")/("Density")=(15)/(7500)m^(3)` `therefore "Hysteresis loss", W=300xx(15)/(7500)xx25xx3600J=5400J` |
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| 4. |
Two tangent galvanometers having coils of the same radius are connected in series. A current flowing in them produces deflections of `60^(@)` and `45^(@)` respectively. The ratio of the number of turns in the coils is |
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Answer» Correct Answer - D In a series current is same. `Rightarrow " " (M_(1))/(M_(2))=(tan theta_(1))/(tan theta_(2)) or (N_(1))/(N_(2))=(tan60^(@))/(tan45^(@))=sqrt3` |
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| 5. |
A bar magnet of pole strength 10A-m is cut into two equal parts breathwise. The pole strength of each magnet isA. 5A-mB. 10A-mC. 15AD. 15A-m |
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Answer» Correct Answer - A When magnet is cut into equal parts, area of cross-section becomes half. Hence pole strength becomes half. |
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| 6. |
A bar magnet of magnetic moment M and pole strength m is broken in two equal part along the magnetic axis . The magnetic moment and pole strength of each part areA. `(M)/(2) , (m)/(2)`B. `(M)/(2)` , mC. `M , (m)/(2)`D. M , m |
| Answer» Correct Answer - A | |
| 7. |
Assertion : When a bar magnet is freely suspended, it points in the north-south direction. Reason : The earth behaves as a magnet with the magnetic field pointing approximately from the geographic south to north.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true and reason is not the correct explanation of assertion.C. If assertion is ture but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - D The magnet align itself in north-south direction, when it is freely suspended because earth behaves as a magnet. |
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| 8. |
The time period of oscillation of a freely suspended bar magnet with usual notations is given byA. `T=2pisqrt(I/(MB_(H)))`B. `T=2pisqrt((MB_(H))/I)`C. `T=sqrt(I/(MB_(H)))`D. `T=2pisqrt(B_(H)/(MI))` |
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Answer» Correct Answer - A Theory based |
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| 9. |
The couple acting on a magnet of length 10cm and pole strength 15A-m, kept in a field of `B=2xx10^(-5)`, at an anlge of `30^(@)` isA. `1.5xx10^(-5)N-m`B. `1.5xx10^(-3)N-m`C. `1.5xx10^(-2)N-m`D. `1.5xx10^(-6)N-m` |
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Answer» As, `tau=MB sin theta=(mxx2l)xx2xx10^(-5)sin30^(@)` `=15xx10xx10^(-2)xx(1)/(2)=1.5xx10^(-5)N-m` |
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| 10. |
A magnet of magnetic moment `M` is situated with its axis along the direction of a magnetic field of strength `B`. The work done in rotating it by an angle of `180^(@)` will beA. `-MB`B. `+MB`C. `0`D. `+2MB` |
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Answer» Correct Answer - D Work done `MB(cos theta_(1)-cos theta_(2))` `theta_(1)=0^(@)` and `theta_(2)=180^(@)` `implies W=MB(cos0-cos180)=2MB` |
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| 11. |
If a bar magnet moment M is freely suspended in a uniform magnetic field of strength field of strength B, then the work done in rotating the magent through an angle `theta` isA. `MB(1-sin theta)`B. `MBsin theta`C. `MB cos theta`D. `MB(1-cos theta)` |
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Answer» Work done is rotating the magnet through an angle `theta` from initial position `(i.e.,theta_(1)=0^(@))` is given by `W=MB(cos theta _(1)-cos theta)=MB(cos 0^(@)-cos theta)=MB(1-costheta)` |
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| 12. |
A bar magnet is held at right angle to a uniform magneitc field. The couple acting on the acting on the magnet is to be halved by rotating it form this position. The angle of rotation isA. `60^(@)`B. `45^(@)`C. `30^(@)`D. `75^(@)` |
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Answer» As, `tau=MB sintheta` Where, `theta=90^(@) Rightarrow tau=MB` `tau_(2)=(1)/(2)tau_(1) Rightarrow MB sintheta=(1)/(2)MB Rightarrow sin theta =(1)/(2) Rightarrow theta=30^(@)` `therefore "Angle of rotation"=90^(@)-30^(@)=60^(@)` |
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| 13. |
The work done in rotating a magnet of magnetic moment `2A-m^(2)` in a magnetic field to opposite direction to the magnetic field, isA. ZeroB. `2xx10^(-2)J`C. `10^(-2)J`D. `10J` |
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Answer» Correct Answer - B `W= MB(1-costheta), where theta=180^(@)` `implies W= 2MB implies W=2xx2xx5xx10^(-3) =2xx10^(-2)J` |
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| 14. |
A short bar magnet pleaced with its axis at `30^(@)` with a uniform external magnetic field of `0.16` Tesla expriences a torque of magnitude `0.032` Joule. The magnetic moment of the bar magnet will beA. `0.23 "Joule"//"Tesla"`B. `0.40 "Joule"//"Tesla"`C. `0.80 "Joule"//"Tesla"`D. zero |
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Answer» Correct Answer - B `tau= MB_(H) sintheta implies 0.032=Mxx0.16xx sin 30^(@)` `implies M=0.4J//tesla` |
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| 15. |
A bar magnet has coercivity `4xx10^(3) Am^(-1)`. It is desired to demagnetise it by inserting it inside a solenoid `12 cm` long and having `60` turns. The current that should be sent through the solenoid isA. `2A`B. `4A`C. `6A`D. `8A` |
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Answer» Correct Answer - D The bar magnet coercively `4xx10^(3) Am^(-1)` i.e., it requires a magnetic intensity `H=4xx10^(3) Am^(-1)` to get demagnetized. Let I be the current carried by solenoid having n number of turns per metre length, then by definition `H=ni`. Here `H=4xx10^(3) Amp turn metre^(-1)` `n=N/l=60/0.12=500 turn metre^(-1)` `implies i=H/n=(4xx10^(3))/500=8.0A` |
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| 16. |
If a diamagnetic substance is brought near north or south pole of a bar magnet, it isA. Attracted by the polesB. Repelled by the polesC. Repelled by the north pole and attracted by the south poleD. Attracted by the north pole and repelled by the south pole |
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Answer» Correct Answer - B Repelled due to induction of similar poles. |
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| 17. |
A short bar magnet has a magnetic moment of `0.48 JT^(-1)`. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equitorial lines (normal bisector) of the magnet. |
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Answer» Here `M = 0.48 JT^(-1), B = ?` d = 10 cm = 0.1 m (a) On the axis of the magnet `B = (mu_(0))/(4pi).(2M)/(d^(3)) = 10^(-7) xx (2 xx 0.48)/((0.1)^(3))` `= 0.96 xx 10^(T)` along S - N direction (b) On the equitorial line of the magnet `B = (mu_(0))/(4pi) xx (M)/(d^(3)) = 10^(-7) xx (0.48)/((0.1)^(3)) = 0.48 xx 10^(-4)T`, along N - S direction. |
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| 18. |
A bar magnet having a magnetic moment of `2xx10^(4) JT^(-1)` is free to rotate in a horizontal plane. A horizontal magnetic field `B=6xx10^(-4)T` exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction `60^(@)` from the field isA. `0.6J`B. `12J`C. `6J`D. `2J` |
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Answer» Correct Answer - C The work done in rotating a magnetic dipole against the torque acting on it, when placed in magnetic field is stored inside it in the form of potential energy. When magnetic dipole is rotated from intial position `theta=theta_(1)` to final position `theta=theta_(2)`, then work done `=MB(costheta_(1)-costheta_(2))` `= MB(1-(1)/(2)) = (2 xx 10^(4) xx 6 xx 10^(-4))/(2) = 6 J` |
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| 19. |
A bar magnet of magnetic moment `1.5 JT^(-1)` lies aligned with the direction of a uniform magnetic field of 0.22 T. magnetic moment : (i) normal to the field direction, (ii) opposite to the field direction ?What is the torque on the magnet in cases (i) and (ii) ? |
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Answer» Torque `r = mB sin theta` (i) Here `theta = 90^(@), r = 1.5 xx 0.22 sin 90^(@) = 0.33 Nm` (ii) Here `theta = 180^(@), r = 1.5 xx 0.22 sin 180^(@) = 0` |
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| 20. |
Which of the following is not correct about relative magnetic permeability `(mu_(r))`?A. It is a dimensionless pure ratio.B. For vacuum medium its value is one.C. For ferromagnetic materials of `mu gt gt 1`D. For paramagnetic materials `mu_(r) gt 1` |
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Answer» Correct Answer - D Relative magnetic permeability `mu_(r)=("magnetic permeability of material"(mu))/("permeability of free space"(mu_(0)))` It is a dimensionless pure ratio and for paramagnetic materials `mu gt 1` |
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| 21. |
Variation of `mu_(r)` (relative permeability) with magnetizing field as shown in figure is for A. Ferromagnetic materialB. Diamagnetic materialC. Paramagnetic materialD. All of these |
| Answer» Correct Answer - A | |
| 22. |
Assertion (A): Relative magnetic permeability has no units and no dimmensions Reason (R ): `mu_(r)=mu//mu_(0)`, where the symbols have their standard meaning.A. If both Assertain and Reason are true and Reason is the correct explanation of AssertainB. If both Assertain and Reason are true but Reason is not correct explantion of AssertainC. If Assertion is true but Reason is falseD. If Assertion is false but Reason in true. |
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Answer» Correct Answer - A Relative magnetic permeability is given by `" " mu_(r)=(mu)/(mu_(0))` It is a dimensionless quantities. |
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| 23. |
Assertion: The poles of magnet cannot be separated by breaking into two pieces. Reason: The magnetic moment will be reduced to half when a magnet is broken into two equal pieces.A. If both Assertain and Reason are true and Reason is the correct explanation of AssertainB. If both Assertain and Reason are true but Reason is not correct explantion of AssertainC. If Assertion is true but Reason is falseD. If Assertion is false but Reason in true. |
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Answer» Correct Answer - B When we break magnet into two pices its poles does not get separated as it again becomes a magnet. Also, when magnet is broken into tiwo equal pieces, its magnetic moments gets reduced to half. |
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| 24. |
Assertion: The poles of magnet cannot be separated by breaking into two pieces. Reason: The magnetic moment will be reduced to half when a magnet is broken into two equal pieces.A. If both the assertion and reason are true and reason is the correct explanation of the assertion.B. If both the assertion and reason are true but the reason is not the correct explanation of assertion.C. If the assertion is true is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - B As we know every atom of a magnet acts as dipole, So poles cannot be separated. When magnet is broken into two equal pieces, magnetic moment of each part will be half of the original magnet. |
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| 25. |
A paramagnetic sample shows a net magnetisation of `8Am^-1` when placed in an external magnetic field of `0*6T` at a temperature of `4K`. When the same sample is placed in an external magnetic field of `0*2T` at a temperature of `16K`, the magnetisation will beA. `(32)/(3)Am^(-1)`B. `(2)/(3)Am ^(-1)`C. `6Am^(-1)`D. `2.4Am^(-1)` |
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Answer» As curie law explains, we can deduce a formula for the relation between magnetic field induction, temperature and magnetisation. i.e., I(magnetisation) `prop(B ("magnetic field induction"))/(t("temperature in kelvin"))` `rArr (I_(2))/(I_(1))=(B_(2))/(B_(1))xx (t_(1))/(t_(2))` Let us suppose, here `I_(1)=8Am^(-1)` `B_(1)=0.6T,t_(1)=4k` `B_(2)=0.2T,t_(2)=16k` `I_(2)=?` `rArr (0.2)/(0.6)xx(4)/(16)=(I_(2))/(8)` `rArr I_(2)=8xx(1)/(12)=(2)/(3)Am^(-1)` |
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| 26. |
A vibration magnetometer is placed at the south pole, then the time period will beA. zeroB. infinityC. same as at magnetic equatorD. same as at any other place on earth |
| Answer» Correct Answer - B | |
| 27. |
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 the above |
| Answer» Correct Answer - B | |
| 28. |
Intensity of magnetisation isA. VectorB. ScalarC. TensorD. Dimensionless vector quantity |
| Answer» Correct Answer - A | |
| 29. |
The arms of a deflection magnetometer in the tan B posittion are placedA. east-westB. north- southC. north-eastD. south- west |
| Answer» Correct Answer - B | |
| 30. |
A short bar magnet placed with its axis at `30^(@)` with an external field of 800 G experiences a torque of 0.016 Nm. The bar magnet is replaced by a solenoid of cross-sectional area `2 xx 10^(-4) m^(2)` and 1000 turns, but of the same magnetic moment. Determine the current flowing through the solenoid. |
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Answer» From Eq. `m_(s) = NIA`. From part (a), `m_(s) = 0.40 A m^(2)` `0.40 = 1000 xx 1 xx 2 xx 10^(-4)` `I = 0.40 xx 10^(4)//(1000 xx 2) = 2A` |
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| 31. |
Figure shows a small magnetised needle P placed at a point O. The arrow shows the direction of magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle Q. (a) In which configuration is the system not in equilibrium? (b) In which configuration is the system in (i) stable and (ii) unstable equilibrium? (c) Which configuration corresponds to the lowest potential energy among all the configurations shown? |
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Answer» Potential energy of the configuration arises due to the potential energy of one dipole (say, Q) in the magnetic field due to other (P). Use the result that the field due to P is given by the expression [Eqs. (5.7) and (5.8)]: `B_(P)=(mu_(0))/(4pi)(m_(r))/(r^(3))` (on the normal bisector) `B_(P)=(mu_(0)2)/(4pi)(m_(r))/(r^(3))` (on the axis) Where `m_(p) ` is the magnetic moment of the dipole P. Equilibrium is stable when `m_(Q)` is parallel to `B_(P)`. and unstable when It is anti-parallel to `B_(P)`. For instance for the configuration `Q_(3)` for which Q is along. the perpendicular bisector of the dipole P. the magnetic moment of Q is parallel to the magnetic field at the position . Hence `Q_(3)` is stable. thus, (a) `PQ_(1)` and `PQ_(2)` (b) (1) `PQ_(3).PQ_(6)` (stable): (II) `PQ_(5).PQ_(4)` (unstable) (c) `PQ_(6)` |
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| 32. |
Assertion: Basic difference between an electric line and magnetic line of force is that former is discontinuous and the latter is continuous or endless. Reason: No electric lines of force exist inside a charged body but magnetic lines do exist inside a magnet.A. If both Assertain and Reason are true and Reason is the correct explanation of AssertainB. If both Assertain and Reason are true but Reason is not correct explantion of AssertainC. If Assertion is true but Reason is falseD. If Assertion is false but Reason in true. |
| Answer» Correct Answer - B | |
| 33. |
What is the basic use of hysteresis curve?A. Paramagnetic materialsB. Diamagnetic materialsC. Ferromagnetic materialsD. All of these |
| Answer» Correct Answer - D | |
| 34. |
A deflection magnetometer is adjusted in the usual way . When a magnet is introduced , the deflection observed is `theta = 60^(@)` and the period of oscillation of the needle in magnetometer is T . When the magnet is removed the time period of oscillation is `T_(0)` . The relation between T and `T_(0)` isA. `2 T^(2) = T_(0)^(2)`B. `(2T^(2))/(sqrt3) = T_(0)^(2)`C. `T = T_(0)`D. `T^(2) = (T_(0)^(2))/(sqrt2)` |
| Answer» Correct Answer - A | |
| 35. |
The magnetic moment of a short bar magnet placed with its magnetic axis at `30^(@)` to an external field of 900 G and experiences a torque of 0.02 N m isA. `0.35Am^(2)`B. `0.44Am^(2)`C. `2.45Am^(2)`D. `1.5Am^(2)` |
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Answer» Correct Answer - B Here, B=900, Gauss=`900xx10^(-4)T` `=9xx10^(2)T` `tau=0.02Nm and theta=30^(@)` `Rightarrow 0.02=9xx10^(-2)xx(1)/(2)xxm` `m=(0.02xx2)/(9xx10^(-2))=0.44Am^(2)` |
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| 36. |
Assertion (A): It is not necessary that every magnet has one north pole and one south pole. Reason (R ): It is a basic fact that magnetic poles occur in pairsA. If both Assertain and Reason are true and Reason is the correct explanation of AssertainB. If both Assertain and Reason are true but Reason is not correct explantion of AssertainC. If Assertion is true but Reason is falseD. If Assertion is false but Reason in true. |
| Answer» Correct Answer - D | |
| 37. |
An iron nail near a bar magnet experiencesA. Only torqueB. Torque and force of attractionC. Only forceD. Torque and force of repulsion |
| Answer» Correct Answer - C | |
| 38. |
Figure shows a small magnetised needle P placed at a point O. The arrow shows the direction of magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle Q. (a) In which configuration is the system not in equilibrium? (b) In which configuration is the system in (i) stable and (ii) unstable equilibrium? (c) Which configuration corresponds to the lowest potential energy among all the configurations shown?A. `PQ_(3)`B. `PQ_(4)`C. `PQ_(5)`D. `PQ_(6)` |
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Answer» Correct Answer - D As potential energy is given as `theta=tan^(-1)((tan delta_(1))/(tan delta_(2)))` Hence, M and B parallel to each other for minimum potential energy. `PO_(6)` configuration provides `theta=0(@), U_(min)=-MB` |
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| 39. |
A magnetic dipole placed in a magnetic field experiences a net force. What can you say about the nature of the magnetic field ? |
| Answer» The nature of the magnetic field is uniform, magnetic dipole (bar magnet) experiences a net force (or torque). | |
| 40. |
Do you find two magnetic field lines intersecting ? Why ? |
| Answer» Two magnetic field lines never intersect. If they intersect, at the point of intersection the field can have two directions. This is impossible. So, two field lines never intersect. | |
| 41. |
What are the units of magnetic moment, magnetic induction and magnetic field ? |
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Answer» The unit of magnetic moment (M) is ampere (M) is ampere - `"meter"^(2) (Am^(2)).` The unit of induction (B) is tesla (T) or N/A - m. The unit of magnetic field (B) is tesla. |
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| 42. |
What are the SI units of magnetic field induction or magnetic flux density?A. teslaB. `weber//meter^(2)`C. newton / ampere-meterD. All of these |
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Answer» Correct Answer - D SI units of magnetic induction is `B=(Units of phi)/(Unit of DeltaA)=(Wb)/(m^(2))=Telsa` Also, SI units of B is `weber/meter^(2)` and Neqton/amp-mtre. |
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| 43. |
(a) What happens if a bar magnet is cut into two pieces (i) transverse to its length (ii) along its length? (b) What happens if an iron bar magnet is melted? Does it retain its magnetism? (c) A magnetised needle in a uniform magnetic field experiences a torque but no net force. However, an iron nail near a bar magnet experiences a force of attration in addition to a torque, explain. (d) Must every magnetic field configuration have a north pole and a south pole? What about the field due to a toroid? (e) Can you think of magnetic field configuration with three poles? (f) Two identical looking iron bars A and B are given, one of which is definitely known to be magnetised. How would one ascertain whether or not both are magnetised? If only one is magnetised how does one ascertain which one? Use nothing else but the bars A and B. |
| Answer» (a) In either case, one gets two magnets, each with a north and south pole. (b) No force if the field is uniform. The iron nail experiences a non- uniform field due to the bar magnet. There is induced magnetic moment in the nail, therefore, it experiences both force and torque. The net force is attractive because the induced south pole (say) in the nail is closer to the north pole of magnet than induced north pole. (c) Not necessarily. True only if the source of the field has a net non- zero magnetic moment. This is not so for a toroid or even for a straight infinite conductor. (d) Try to bring different ends of the bars closer. A repulsive force in some situation establishes that both are magnetised. If it is always attractive, then one of them is not magnetised. In a bar magnet the intensity of the magnetic field is the strongest at the two ends (poles) and weakest at the central region. This fact may be used to determine whether A or B is the magnet. In this case, to see which one of the two bars is a magnet, pick up one, (say, A) and lower one of its ends, first on one of the ends of the other (say, B), and then on the middle of B. If you notice that in the middle of B, A experiences no force, then B is magnetised. If you do not notice any change from the end to the middle of B, then A is magnetised | |
| 44. |
A magnet is placed in iron poweder and the taken out , them maximum iron powder is atA. some distannce away from north poleB. some distance away from north poleC. the middle of the magnetD. the end of the magnet |
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Answer» Correct Answer - D Magnetic pole strength is stronger at end parto of magnet. So maximum iron powder is collected at the end point of magnet. |
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| 45. |
A dipole of magnetic moment `vec(m)=30hatjA m^(2)` is placed along the y-axis in a uniform magnetic field `vec(B)= (2hat i +5hatj)T`. The torque acting on it isA. `-40hatk Nm`B. `-50hatk Nm`C. `-60hatk Nm`D. `-70hatk Nm` |
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Answer» Correct Answer - C `"Here", vec(m)=30hatjAm^(2) and vec(B)=(2hati+5hatj)T` `"Since", vec(tau)=vec(m)xxvec(B)` `=30hatjxx(2hati+5hatj)=60hatjxxhati+150hatjxx 60(-hatk)+150xx0` `=-60hatkNm [therefore hatj xxhatj=-hatkand hatjxxhatj=0]` |
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| 46. |
A short bar magnet is kept in uniform magnetic field of `0.16` T such that dipole moment vector is at angle `30^(@)` with the field . The torque acting on the bar magnet is 0.032 N m . What is the magnitude of magnetic dipole moment of the magnet ? |
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Answer» `tau = MB sin theta` `0.032 = M xx 0.16 xx (1)/(2)` `0.064 = M xx 0.16` `M = (6.4)/(16) = 0.4 Am^(2)` |
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| 47. |
A rod of ferromegnetic materical with dimensions `10cmxx0.5cmxx2cm` is placed in a magnetising field of intensity `2xx10^(5)A//m`. The magnetic moment produced due it is `6` `amop-m^(2)`. The value of magnetic induction will be------`10^(-2)T`.A. 0.358TB. 0.54TC. 6.28TD. 2.519T |
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Answer» Correct Answer - B `"Here", V=10xx0.5xx0.2cm^(3)=10^(-6)m^(3)` `H=0.5xx10^(4)Am^(-1)=5xx10^(3)Am^(-1)` `M=5Am^(-2), B=?` `I=(M)/(A)=(5)/(10^(-6))=5xx10^(6)Am^(-1)` `B=mu_(0)(H+I)` `B=mu_(0)(H+I)` `=4pixx10^(-7)(5xx10^(3)+5xx10^(6))` `=4pixx10^(-7)xx5xx10^(6)(10^(-3)+1)=6.28T` |
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| 48. |
A small bar magnet of moment M is placed in a uniform field H. If magnet makes an angle of `30^(@)` with field, the torque acting on the magnet isA. `MH`B. `(MH)/2`C. `(MH)/3`D. `(MH)/4` |
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Answer» Correct Answer - B `tau=MHsintheta=MHsin30^(@)=(MH)/2` |
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| 49. |
A bar magnet has a magnetic moment of 200 A m? The magnet is suspended in a magnetic field of 0.30 `"N A"^(-1)" m"^(-1)`. The torque required to rotate the magnet from its equilibrium position through an angle of `30^(@)`, will beA. 30 N mB. `30sqrt(30)` N mC. 60 N mD. `60sqrt(3)` N m |
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Answer» Correct Answer - A Torque experienced by a magnet suspended in a uniform magnetic field `B` is given by `tau=MBsintheta` Here, `M=200"A m"^(2)`, B= 0.30 N `A^(-1) m^(-1)` and `theta= 30^(@)` `thereforetau=200xx0.30xxsin30^(@)` `tau=30"N m"` |
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| 50. |
The electron in hydrogen atom moves with a speed of `2.2xx10^(6)m//s` in an orbit of radius `5.3xx10^(-11)cm`. Find the magnetic moment of the orbiting electron. |
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Answer» Frequency of revolution, `f=(v)/(2pir)` The moving charge is equilvalent to a current loop, given by `" " I=fxxe or I=(ev)/(2pir)` If A be the area of the orbit, then the magnetic moment of the orbiting electrons is `" " M=IA=((ev)/(2pir))(pir^(2))=(evr)/(2)` Putting the value, we get `" " M=((1.6xx10^(-19))(2.2xx10^(6))(5.3xx10^(-11)xx10^(-2)))/(2)` `" " M==9.3xx10^(-26)A-m^(2)` |
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