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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.
| 701. |
In Fig. a coil B is connected to a low voltage bulb L and placed parallel to another coil A, as shown. Explain the following observations : (i) bulb lights (ii) gets dimmer if the coil B is moved upwards. |
| Answer» (i) Bulb lights because of emf induced in coil B due to mutual induction between A and B (ii) On moving coil B upwards, coeff, of mutual inductance (M) decreases. Induced emf (e) decreases. Bulb gets dimmer. | |
| 702. |
A coil A is connected to a voltmeter V and the other coil B to an alterniting current sourecs, Fig. If a large copper sheet is placed between the two coils, how does the induced emf in coil A change due to current in coil B ? |
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Answer» In the source of copper sheet, emf will be induced in coil A due to mutual induction between the two coils A and B. Voltemter in coil A will show deflection as per the magnetude of emf induced. When a copper sheet is placed between the two coils eddy currents are set up in the sheet due to change in magnetic flux linked with C. Magnetic flux linked with coil A, due to col B and due to opposing eddy currents in C will decrease. Therefore, rate of change of magnetic flux in A will decrease. Hence emf induced in A will decrease. |
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| 703. |
Two long parallel horizontal rails a, a distance d aprt and each having a risistance `lambda` per unit length are joing at one end by a resistance R. A perfectly conduction rod MN of mass m is free to slide along the rails without friction (see figure). There is a uniform magnetic field of induction B normal to the plane of the paper and directed into the paper. A variable force F is applied to the rod MN such that, as the rod moves a constant current flows through R. (i) Find the velocity of the rod and the applied force F as function of the distance x of the rod from R. (ii) What fraction of the work done per second by F is converted into heat? |
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Answer» Correct Answer - `[(a) (i(r+2 lamdax))/(Bd),m(( 2i^(2)lamda(R+2lamdax))/(B^(2)d^(2)))(b)(B^(3)d^(3))/(2milamda(R+2lamdax))]` |
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| 704. |
The time constant of an inductance coil is `2 xx 10^(-3) s`. When a `90 Omega` resistance is joined in series, the same constant becomes `0.5 xx 10^(-3) s`. The inductance and resistance of the coil areA. 30mH,`30Omega`B. 30mH,`60Omega`C. 60mH,`30Omega`D. 60mH,`60Omega` |
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Answer» Correct Answer - C Time constant, `tau=(L)/(R)=2.0xx10^(-3)s` and `(L)/(R+90)=0.5xx10^(-3)s` Solving these two equations, we get L = 60 mH and R = 30 `Omega` |
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| 705. |
The magnitude of induced e.m.f. in a conductor, depends uponA. resistance of the conductorB. strength of the magnetic fieldC. orientation of the conductorD. rate of change of flux linkage with the conductor |
| Answer» Correct Answer - D | |
| 706. |
A magnetic field B is confined to a region `r le` a and points out of the paper (the z-axis), r = 0 being the centre of the cicular region. A charged ring (charge = Q) of radius b,`bgt a` and mass m lie in the x-y plane with its centre at origin. The ring is free to rotate and is at rest. The magnetic field is brought to zero in time `Delta t`. Find the angular velocity `omega` of the ring after the field vanishes. |
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Answer» If E is the electric field generated aroung the charged ring of radius b, then as `e = (d phi)/(dt)` `oint E.dl = (B. pi a^(2))/(Delta t)` `E. 2 pi b = (B pi a^(2))/(Delta t)` or `Eb = (B a^(2))/(2(Delta t))` Torque acting on the ring `tau = b xx force = b xx Q xx E` Using (i), `tau = (Q B a^(2))/(2 (Delta t))` If `Delta L` is change in angluar momentum of the charged ring, then as `tau = (Delta L)/(Delta t) = (L_(2) - L_(1))/(Delta t) :. L_(2) - L_(1) = tau (Delta t) = (Q B a^(2) (Delta t))/(2 (Delta t)) = (Q B a^(2))/(2)` As initial angular momentum, `L_(1) = 0 :. L_(2) = (Q B a^(2))/(2) = I omega = (mb^(2)) omega` `omega = (Q B a^(2))/(2 mb^(2))` This is the angular velocity of the ring after the field vanishes. |
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| 707. |
The direction of induced currnet in the coils A and B in the situation shown in the figure is A. p to q in coil A and x to y in coil BB. q to p in coil A and y to x in coil BC. p ot q in coil A and y to x in coil BD. q to p in coil A and y to x in coil B |
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Answer» Correct Answer - B In coil, A south pole is developed at q and in coil B also south pole is developed at x. therefore, induced current in coil A will be from q to p and induced current in the coil B will be from x to y. |
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| 708. |
A magnetic field B is confined to a region `r le` a and points out of the paper (the z-axis), r = 0 being the centre of the cicular region. A charged ring (charge = Q) of radius b,bgt a and mass m lie in the x-y plane with its centre at origin. The ring is free to rotate and is at rest. The magnetic field is brougth to zero in time `Delta t`. Find the angular velocity `omega or the ring after the field vanishes.A. `(qBa^(2))/(2mb)`B. `(qBa)/(2mb^(2))`C. `(2b^(2))/(qBa^(2))`D. `(qb^(2))/(2Ba^(2))` |
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Answer» Correct Answer - D Let E is the electric field generated around the charged ring of radius of b, then, `epsi=(dphi)/(dt)` `ointvec(E)*dvec(l)=(Bpia^(2))/(Deltat)` `or Eb=(Ba^(2))/(2(Deltat))" n "......(i)` Torque acting on the ring `tau=bxx"forces"=bqE` `=(qBa^(2))/(2(Deltat)) " "` [ Using (i)] If `DeltaL` is change in angular momentum of the chorged ring, then, `tau=(DeltaL)/(Deltat)=(L_(2)-L_(1))/(Deltat)` `:. L_(2)-L_(1)=tau(Deltat)` `=(qBa^(2)Deltat)/(2Deltat)=(qBa^(2))/(2)` As initial angular momentum, `L_(1)=0` `:. L_(2)=(qBa^(2))/(2)=I omega=mb^(2)omega :. omega=(qBa^(2))/(2mb^(2))` |
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| 709. |
A metal ring is held horizontally and bar magnet is dropped through the ring with its length along the axis of the ring. The acceleration of the falling magnetA. equal to gB. less than gC. more than gD. depends of on the diameter of the ring length of magnet |
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Answer» Correct Answer - B Acceleation of the magenet will not be equal to g. It will be less than g. This is because, as the magnet falls, amount of mgnetic flux linked with the ring changes. An induced emf is developed in the ring which opposes the downward motion of the magnet. |
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| 710. |
A heart pacing device consists of a coil of `50` turns `&` radius `1 mm` just inside the body with a coil of `1000` turns `&` radius `2 cm` placed concentrically and co axially just outside the body.Calculate the average induced `EMF` in the internal coil, if a current of `1A` in the external coil collapses in `10` milliseconds. |
| Answer» Correct Answer - C::D | |
| 711. |
A flate circular coil of `10 cm` radius has `200` turns of wire the coil is connected to a capacitor of `20 muF` placed in a uniform magnetic field whise induction decreases at a rate of `0.01 Ts^(-1)` Find the charge on capacitorA. `0.51 muC`B. `0.75 muC`C. `0.92 muC`D. `1.25 muC` |
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Answer» Correct Answer - D `Q=Ec,E=(dphi)/(dt)` `Q=(dphi)/(dt).C=(d)/(dt)(NAB)C=NAC(dB)/(dt)` `Q=200xxpixx0.1^(2)xx10^(-2)xx20xx10^(-6)` `=1.25xx10^(-6)C=1.25muC` |
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| 712. |
A solenoid of lenght 30 cm with 10 turns per centimetre and area of cross-section `40cm^(2)` completely surrounds another co-axial solenoid of same length, area of cross-section `20cm^(2)` with 40 turns per centimetre. The mutual inductance of the system isA. 10 HB. 8 HC. 3 mHD. 30 mH |
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Answer» Correct Answer - C Mutual inductance of the system, `M=mu_(0)n_(1)n_(2)A_(2)l` where `A_(2)` is the area of inner solenoid. `n_(1)=10 cm^(-1)=1000 m^(-1)` `n_(2)=40 cm^(-1)=4000 m^(-1)` `l=30 cm=30xx10^(-2)m` `A_(2)=20cm^(2)=20xx10^(-4)m^(2)` `:.M=4pixx10^(-7)xx1000xx4000xx20xx10^(-4)xx30xx10^(-2)` `=301.44xx10^(-5)H=3mH` |
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| 713. |
A thin circular ring of area `A` is perpendicular to uniform magnetic field of induction `B`. `A` small cut is made in the ring and a galavanometer is connected across the ends such that the total resistance of circuit is `R`. When the ring is suddenly sqeezed to zero area, the charge flowing through galvanometer is:A. `(BR)/(A)`B. `(AB)/(R )`C. `ABR`D. `(B^(2)A)/(R^(2)` |
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Answer» Correct Answer - B `Q=(Ndeltaphi)/(R )=(phi_(2)-phi_(1))/(R )=(BA-0)/(R )=(BA)/(R )` |
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| 714. |
In electromagnetic induction, the induced charge in a coil is independent ofA. change in the fluxB. timeC. resistance in the circuiteD. none of the above |
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Answer» Correct Answer - B We know that `e=(dvarphi)/(dt)` but e=iR and `i=(dq)/(dt)implies(dq)/(dt) R=dq=(dvarphi)/(R )` |
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| 715. |
In a long staright solenoid with cross-sectional radius `a` and number of turns per unit length `n` a current varies with a constant velocity` dotI A//s`. Find the magntidue of the eddy current field strength as a function of the distance `r` from the solenoid axis. Draw the appoximate plot of this function. |
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Answer» Correct Answer - For `r lt a, (1)/(2)rmu_(0) " nl and for " r gt a, (1)/(2)(a^(2)mu_(0)nI)/(r )` |
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| 716. |
Whenever the magnet flux linked with a coil changes, then is an induced emf in the circuit. This emf lastsA. for everB. for a long timeC. for a short timeD. so long as the change in flux take place |
| Answer» Correct Answer - D | |
| 717. |
The magnitude of induced emf during electromagnetic induction, is directly proportional toA. electric fluxB. magnetic fluxC. magnetic fieldD. electric field |
| Answer» Correct Answer - B | |
| 718. |
If the rate of change of current in primary coil is doubled, the induced emf in secondary coil becomeA. halfB. sameC. doubleD. 4 times |
| Answer» Correct Answer - C | |
| 719. |
The mutual inductance depends uponA. number of turns of a coilB. geometrical properties of a coilC. medium between the two coilD. all of these |
| Answer» Correct Answer - D | |
| 720. |
The displacement current isA. `I_(D)=epsi_(0)(dphi_(E))/(dt)`B. `I_(D)=(1)/(epsi_(0))(dphi_(E))/(dt)`C. `I_(D)=epsi_(0)(d^(2)phi_(E))/(dt)`D. `I_(D)=(1)/(epsi_(0))(d^(2)phi_(E))/(dt)` |
| Answer» Correct Answer - A | |
| 721. |
Assertion : Sensitive electrical instruments should not be placed in the vicinity of an electromagent. Reason : Electromagnet can damage the instruments. |
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Answer» Correct Answer - a Senstive electrical instrument instrum in the vicinity of electromeagent can be damaged due to the inducedd emfs and the resulting currens when the electrognets is turned nor or off. |
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| 722. |
the wave of electric and magnetic field both varying in space and time isA. matter waveB. sound waveC. longitudinal waveD. electromagnetic wave |
| Answer» Correct Answer - D | |
| 723. |
The core of a transformer is laminated to reduceA. eddy currentB. hysteresisC. resistance in windingD. none of these |
| Answer» Correct Answer - A | |
| 724. |
Assertion : An electric motor converts electical energy to mechanical energy. Reason : The working of the motor is based on mutualn induction. |
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Answer» Correct Answer - c A dc motor converts electric energy into mechanicla energy. The principle on which it is base is not mutual inductio. A current carrying coil, in a magnetic field experiences a torque. This tourque rotates the coil. The dc motor is based on the interaction of current and magentic field. |
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| 725. |
Assertion : An ac generator is based on the self inductance of the coil. Reason : Self inductance involves two coils. |
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Answer» Correct Answer - d An ac generator is based on the phenomenon of electtromagnetic induction. Self inductance involves a single coil |
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| 726. |
The back emf in a DC motor is maximum when, |
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Answer» Correct Answer - d Back emf is actually the induced emf produced in motor. When motor is just swotched on, the magnitude of back emf is quite small due to the low speed of the motor. But the magnitude of back emf is very large at break (when electric suply fails suddenly, while the motor is running) |
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| 727. |
The coefficient of mutual induction between two circuits is equal to the, emf produced in one circuit when the current in the second circuit isA. Kept steady at 1 ampereB. Cut-off at 1 ampere levelC. Changed at the rate of 1 A/sD. Changed from 1 A/s to 2 A/s |
| Answer» Correct Answer - C | |
| 728. |
Large transformers, when used for some time, become hot and are cooled by circulating oil. The heating of transformer is due toA. heating effect of current aloneB. hysteresis loss aloneC. both the hysteresis loss and heating effect of currentD. none of these |
| Answer» Correct Answer - C | |
| 729. |
An `LR` series circuit has `L=1 H` and `R=1 Omega`.It is connected across an `emf` of `2 V`.The maximum rate at which energy is stored in the magnetic field is: A. The maximum rate at which energy is stored in the magnetic field is 1W.B. The maximum rate at which energy is stored in the magnetic field is 2W.C. The current at that instant is 1AD. The current at that instant is 2A. |
| Answer» Correct Answer - A::C | |
| 730. |
An `LR` series circuit has `L=1 H` and `R=1 Omega`.It is connected across an `emf` of `2 V`.The maximum rate at which energy is stored in the magnetic field is: A. The maximum rate at which energy is stored in the magnetic field is `1W`B. The maximum rate at which energy is stored in the magnetic field is `2W`C. The current at that instant is `1 A`D. The current at that instant is `2 A` |
| Answer» Correct Answer - A::C | |
| 731. |
Core of transformer is made upA. soft ironB. steelC. ironD. alnico |
| Answer» Correct Answer - A | |
| 732. |
An LR circuit with a battery is connected at t=0. Which of the following quantities is not zero just after the connection?A. current in the circuitB. magnetic field energy in the inductorC. power delivered by the batteryD. `emf` induced in the inductor |
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Answer» Correct Answer - A::B::C `EMF` induced =`-L(di)/(dt) != 0`, rest quantities are zero. |
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| 733. |
An LR series circuit with a battery is connected at t=0. Which of the following quantities are zero just after the connection?A. current in the circuitB. magnetic field energy in the inductorC. power delivered by the batteryD. emf induced in the inductor. |
| Answer» Correct Answer - A::B::C | |
| 734. |
What is magnetic flux linked with a coil or N turns or are cross section A held with its plane parallel to the field? |
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Answer» As plane of area of coil is parallel to the field, normal to the area makes an angle of `90^(@)` with the field. Therefore, `phi = N BA cos 90^(@) =` Zero |
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| 735. |
A square of side x m lies in the x-y plane in a region, when the magntic field is given by `vecB=B_(0)(3hati+4hatj+5hatk)` T, where `B_(0)` is constant. The magnitude of flux passing through the square isA. `5B_(0)x^(2)Wb`B. `3B_(0)x^(3)Wb`C. `2B_(0)x^(2)Wb`D. `B_(0)x^(2)Wb` |
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Answer» Correct Answer - A Here, `vecA=x^(2)hatkm^(2)and vecB=B_(0)(3hati+4hatj+5hatk)T` AS `phi=vecB.vecA=B_(0)(3hati+4hatj+5hatk).x^(2)hatk` `:.phi=5B_(0)x^(2)Wb` |
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| 736. |
What e.m.f. will be induced in a 10 inductor in which current changes from 10 A to 7 A in `9xx10^(-2) s` ? |
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Answer» Here, `e = ?, L = 10 H, I_(1) = 10 A, I_(2) = 7 A, dt = 9 xx 10^(-2) s` As `e = (L dl)/(dt) = (-L(I_(2)-I_(1)))/(dt) = (-10(7 - 10))/(9 xx 10^(-2)) = 333.3` volt. |
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| 737. |
A condenser of capacitor `0.144 pF` is used in a transmitte to transmit at wavelength `lambda`. If inductace of `1//pi^(2) mH` is used for resonance, what is the value of `lambda` (in m) ? |
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Answer» Correct Answer - 7 Here,`C = 0.144 pF = 0.144 xx 10^(-12) F` `L = (1)/(pi^(2)) mH = (10^(-3))/(pi^(2)) H, v = (1)/(2 pi sqrt(LC))` `lambda = (upsilon)/(v) = upsilon . 2 pi sqrt(LC)` `=3 xx 10^(8) xx 2 pi sqrt((10^(3))/(pi^(2)) xx 0.144 xx 10^(-12))` `lambda = 3 xx 10^(8) xx 2 pi xx (1.2 xx 10^(-8))/(pi) = 7.2 m` The correct answer is `7 m` |
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| 738. |
Which of the primary and secondary of a step down transformer has a larger number of turns ? |
| Answer» Primary coil in a step down transformer has larger number of turns. | |
| 739. |
Same as problem `4` except the coil `A` is made to rotate about a vertical axis in the plane of the coil (Figure). No currents flows in `B` if `A` is at rest. The current in coil `A`, when the current in `B` (at `t = 0`) is counterclockwise and the coil `A` is as shown at this instant, `t = 0`, is A. constant current clockwise.B. varying current clockwise.C. varying current clockwise.D. constant current counterclockwise. |
| Answer» Correct Answer - a | |
| 740. |
The instantaneous current from an a.c. source is `I = 6 sin 314 t`. What is the r.m.s value of the current ? |
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Answer» Compare the given eqn. with the standed from `I = I_(0) sin omega t` `I_(0) = 6 A, I_(rms) = (I_(0))/(sqrt2) = (6)/(sqrt2). (sqrt 2)/(sqrt 2) = (6)/(2) xx 1.414 = 4.242 A` |
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| 741. |
Does a transformer change the frequency of a.c. ? |
| Answer» No, the frequency cannot be changed by a transformer. | |
| 742. |
How can you improve the quality factor of series resonance circuit ? |
| Answer» To improve the quality factor of series resonance circuit, ohmic resistance of the circuit should be made as small as possible | |
| 743. |
What is the function of oil in a transformer ? |
| Answer» It provides insulation as well as cooling. | |
| 744. |
What is iron in a transformer ? |
| Answer» Iron loss is loss of energy in the form of heat in iron core of a transformer. | |
| 745. |
What is the frequency of rotation of armature coil in AC generators? |
| Answer» 50 Hz in India and UK. It is 60 Hz in the USA. | |
| 746. |
A sheet of relative permeability 100 is placed inside the solenoid, so as to fill the space. What will be the energy stored inside the solenoid? |
| Answer» `(50B^(2))/(mu_(0))` | |
| 747. |
If self induction of an air core solenoid increases from 0.04 mH to 16 mH on introducing an iron core into it, what is relative magneitc permeability the core ? |
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Answer» As `L_(0) = (mu_(0) N^(2) A)/(l)` and `L = (mu N^(2) A)/(l)` `:. mu_(r) = (mu)/(mu_(0)) = (L)/(L_(0)) = (16)/(0.04) = 400` |
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| 748. |
A `50cm` long solenoid having `500` turns and radius `2cm` is wound on an iron core of relative permeability `800` . What will be the average emf induced in the solenoid if the current in it changes from `0` to `2A` in `0.05` sec? |
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Answer» For solenoid `L=(mu_(0)mu_(r)N^(2)pir^(2))/(l)` `=((4pixx10^(-7))(800)(500)^(2)pi(0.02)^(2))/(0.5)` `=0.64H` `bar(e)=-L(Deltai)/(Deltat)=-(0.64(2-0))/(0.05)=25.6V` |
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| 749. |
Two electric circuits are placed near one another. Each circuit has self inductance Must there be a mutual inductance. |
| Answer» Correct Answer - Yes | |
| 750. |
Find mutual inductance if we have a square loop of side r instead of a small circular loop of side r at the centre of bigger circular loop. |
| Answer» `(mu_(0)r^(2))/(2R)` | |