624 + Interview Questions in MAGNETIC-EFFECT-OF-CURRENT IN GENERAL AWARENESS Page 5 InterviewSolution

201.	A galvanometer with resistance `1000Omega` gives full scale deflection at `01mA`. What value of resistance should be added to `1000Omega` to increase its current range `10A`.A. `001Omega` in seriesB. `001Omega` in parallelC. `010Omega` in seriesD. `0*10Omega` in parallel
Answer» Correct Answer - B Here, `G=1000Omega`, `I_g=01mA=10^-4A`, `I=10A` `S=(I_gG)/(I-I_g)=(10^-4xx1000)/(10-10^-4)` `=001A` in parallel.

Discussion

202.	A galvanometer coil has a resistance of `12Omega` and the meter shows full scale deflection for a current of `3mA`. How will you convert the meter into a voltmeter of range 0 to 18V?
Answer» Here, `G=12Omega`, `I_g=3mA=3xx10^-3A`, `V=18V`, `R=?` From, `R=(V)/(I_g)-G=(18)/(3xx10^-3)-12=6000-12=5988Omega` This resistance is to be connected in series with the galvanometer.

Discussion

203.	A galvanometer with a coil of resistance `12Omega` shows a full scale deflection for a current of `25mA`. How will you convert it into a voltmeter of range `75V`? Also, find the total resistance of voltmeter formed.
Answer» A galvometer can be converted into a voltmeter of given range by connecting a suitable resistance R in series with galvanometer, which is given by, `R=V/I_g-G=(75)/(25xx10^-3)-12=2988Omega` Total resistance of voltmeter formed `=R+G=2988+12=3000Omega`

Discussion

204.	When a current of `5mA` is passed through a galvanometer having a coil of resistance `15Omega`, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into to voltmeter of range `0-10V` isA. `2535xx10^3Omega`B. `4005xx10^3Omega`C. `1985xx10^3Omega`D. `2045xx10^3Omega`
Answer» Here, `I_g=5xx10^-3A`, `G=15Omega`, `V=10V`, `R=V/I_g-G=(10)/(5xx10^-3)-15` `=2000-15=1*985xx10^3Omega`

Discussion

205.	A galvanometer coil has a resistance of `15Omega` and the meter shows full scale deflection for a current of `4mA`. How will you convert the meter into an ammeter of range 0 to 6A?
Answer» Here, `G=15Omega`, `I_g=4mA=4xx10^-3A`, `S=?`, `I=6A` From `S=(I_g.G)/(I-I_g)=(4xx10^-3xx15)/(6-4xx10^-3)=(6xx10^-2)/(5.996)=10^-2ohm` This low resistance, called shunt resistance, is to be connected in parallel with the galvanometer.

Discussion

206.	In a hydrogen atom, the electron moves in an orbit of radius `0*5 Å`, making `10^(16)rps`. Calculate the magnetic moment associated with the orbital motion of electron.
Answer» Here, `r=05Å=05xx10^(-10)m`, `v=10^(16)rps, M=?` Charge on electron, `e=16xx10^(-19)C` Equivalent current, `I=e/T=ev` Area of the orbit, `A=pir^2` As `M=IA` `:. M=ev(pir^2)=16xx10^(-19)xx10^(16)xx314xx(05xx10^(-10))^2` `=1*256xx10^(-23)Am^2`

Discussion

207.	Two long straight parallel conductors separated by a distance of `0.5m` carry currents of `5A` and `8A` in the same direction. The force per unit length experienced by each other isA. `1.6xx10^(-5)N` (attractive)B. `1.6xx10^(-5)N` (repulsive)C. `16xx10^(-5)N` (attractive)D. `16xx10^(-5)N` (repulsive)
Answer» Correct Answer - A

Discussion

208.	Two identical current carrying coaxial loops, carry current I in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as C,A. `oint vecB.dvecl=+-2mu_0I`B. the value of `oint_CvecB.dvecl` is independent of sense of CC. there may be a point on C where `vecB` and `dvecl` are perpendicularD. `vecB` vanishes everywhere on C
Answer» Correct Answer - B::C Consider a simple amperian loop passing once through both the identical current carrying coaxial loops. (i) According to Ampere circuital law, `oint_CvecB.dvecl=mu_0(I-I)=0`. Hence, option (a) is wrong. (ii) As `oint_CvecB.dvecl=0`, therefore `oint_CvecB.dvecl` is independent of sense of C. Thus option (b) is correct. (iii) There will be a point on loop C, lying at the axis of two loops A and B, where `vecB` and `dvecl` are perpendicular to each other. Thus option(c) is correct. (iv) The value of `vecB` does not vanish on various points of C. Thus option (d) is wrong.

Discussion

209.	A long wire bent as shown in fig. carries current I. If the radius of the semicircular portion is a, the magnetic field at centre C is A. `(mu_0I)/(4r)`B. `(mu_0I)/(4pir)sqrt(pi^2+4)`C. `(mu_0I)/(4pi)+(mu_0I)/(2pir)`D. `(mu_0I)/(4pir)sqrt(pi^2-4)`
Answer» Correct Answer - B Total magnetic field induction at P is `vecB=vecB_(ab)+vecB_(bc)+vecB_(cd)` `=(mu_0)/(4pi)I/r(-hatj)+(mu_0)/(4pi)(piI)/(r)(-hatk)+(mu_0)/(4pi)I/r(-hatj)` `=(mu_0)/(4pi)(2I)/(r)(-hatj)+(mu_0)/(4pi)(piI)/(r)(-hatk)` `:. \|vecB\|=[((mu_0)/(4pi)(2I)/(r))^2+((mu_0)/(4pi)(piI)/(r))^2]^(1//2)` `=(mu_0)/(4pi)I/rsqrt(4+pi^2)`

Discussion

210.	A stream of protons is moving parallel to a stream of electrons. Will the two stream tend to come closer or move apart?
Answer» The behaviour of the two streams of protons and electrons coming closer or moving apart depends on their speed. If they have large speeds, they will move apart because the magnetic interaction is more repulsive than electrostatic attraction. If they have small speeds, they will come closer to each other as electrostatic attraction between them is stronger than magnetic repulsion between them.

Discussion

211.	Assertion : If two long wires, hanging freely are connected to a battery in series, they come closer to each other. Reason : Force of attraction acts between the two wires carrying current.A. If both assertion and reason are true and the reason is the correctexplanation of the assertion.B. If both assertion and reason are true but reason is not the correctexplanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false.
Answer» Correct Answer - D

Discussion

212.	Assertion: If two long wires, hanging freely are connected to a battery in series, they come closer to each other. Reason: Force of attraction acts between the two wires carrying currents.A. both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.B. both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both Assertion and Reason are false.
Answer» Correct Answer - D When two long parallel wires, are connected to a battery in series. They carry currents in opposite directions, hence they repel each other.

Discussion

213.	Two long wires are hanging freely. They are joined first in parallel and then in series and then are connected with a battery. In both cases, which type of force acts between the two wiresA. Attraction force when in parallel and repulsion force when in seriesB. Repulsion force when in parallel and attraction force when in seriesC. Repulsion force in both casesD. Attraction force in both cases
Answer» Correct Answer - A

Discussion

214.	Two long parallel wires are at a distance ` 2d` apart. They carry steady equal currents flowing out of the plane of the paper , as shown. The variation of the magnetic field `B` along the line `XX` is given byA. B. C. D.
Answer» Correct Answer - (B) (ii) (a) Field close to wire will tends to OO (b) Field at cetre will be zero (c ) One either side of wire field will be oppo site so option `(B)` will be correct

Discussion

215.	Two wires are held perpendicular to the plane of paper and are 5 m apart. They carry currents of 2.5 A and 5 A in same direction. Then, the magnetic field strength (B) at a point midway between the wires will beA. `(mu_(0))/(4pi)T`B. `(mu_(0))/(2pi)T`C. `(3mu_(0))/(2pi)T`D. `(3mu_(0))/(4pi)T`
Answer» Correct Answer - b

Discussion

216.	Two straight parallel wires, both carrying `10` ampere in the same direction attract each other with a force of `1xx10^(-3)N`. If both currents are doubled, the force of attraction will beA. `1xx10^(-3)N`B. `2xx10^(-3)N`C. `4xx10^(-3)N`D. `0.25xx10^(-3)N`
Answer» Correct Answer - C

Discussion

217.	Three long straight and parallel wires carrying currents are arranged as shown in the figure. The wire `C` which carreis a current of `5.0 amp` is so placed that it experiences no force. The distance of wire `C` from `D` is then A. `9cm`B. `7cm`C. `5cm`D. `3cm`
Answer» Correct Answer - A

Discussion

218.	A circular loop has a radius of `5 cm` and it is carrying a current of `0.1 amp`. It magneitc moment isA. `1.32xx10^(-4)amp-m^(2)`B. `2.62xx10^(-4)amp-m^(2)`C. `5.25xx10^(-4)amp-m^(2)`D. `7.85xx10^(-4)amp-m^(2)`
Answer» Correct Answer - D

Discussion

219.	A long straight wire carries a current of `pi amp`. The magnetic field due to it will be `5xx10^(-5)"weber"//m^(2)` at what distance from the wire [`mu_(0)=` permeability of air]A. `10^(4)mu_(0)` metreB. `(10^(4))/(mu_(0))` metreC. `10^(6)mu_(0)` metreD. `(10^(6))/(mu_(0))` metre
Answer» Correct Answer - A

Discussion

220.	`A, B` and `C` are parallel conductors of equal length carrying currents `I, I` and `2I` respectively. Distance between `A` and `B` is `x`. Distance between `B` and `C` is also `x`. `F_(1)` is the force exerted by `B` on `A` and `F_(2)` is the force exerted by `B` on `C` choose the correct answer A. `F_(1)=2F_(2)`B. `F_(2)=2F_(1)`C. `F_(1)=F_(2)`D. `F_(1)=-F_(2)`
Answer» Correct Answer - D

Discussion

221.	The two linear parallel conductors carrying currents in the opposite direction…………………each other.
Answer» Correct Answer - repel

Discussion

222.	A straight wire carrying a cureent `i_(1)` amp runs along the axis of a circular current `i_(2)` amp. Then the force of interaction between the two current carrying conductors isA. `oo`B. Zero at the centre of loopC. `(mu_(0))/(4pi)(2i_(1)i_(2))/rN//m`D. `(2i_(1)i_(2))/r N//m`
Answer» Correct Answer - B

Discussion

223.	`A` and `B` are two concentric circular conductors of centre `O` and carrying currents `i_(1)` and `i_(2)` as shown in the adjacent figure. If ratio of their radii is `1:2` and ratio of the flux densities at `O` due to `A` and `B` is `1:3`, then the value of `i_(1)//i_(2)` is A. `1/6`B. `1/4`C. `1/3`D. `1/2`
Answer» Correct Answer - A

Discussion

224.	An arbitrary shaped closed coil is made of a wire of length `L` and a current `I` ampere is flowing in it. If the plane of the coil is perpendicular to megnetic field `vecB`, the force on the coil isA. ZeroB. `lBL`C. `2lBL`D. `1/2 IBL`
Answer» Correct Answer - A

Discussion

225.	Two long conductors, separated by a distance `d` carry current `I_(1) and I_(2)` in the same direction . They exert a force `F` on each other. Now the current in one of them is increased to two times and its direction is reversed . The distance is also increased to `3d`. The new value of the force between them isA. `-2F`B. `F//3`C. `2F/3`D. `-F//3`
Answer» Correct Answer - C

Discussion

226.	A unifrom conducting wire `ABC` has a mass of `10g`. A current of `2 A` flows through it. The wire is kept in a unifrom magnetic field `B = 2T`. The accleration of the wire will be A. ZeroB. `12 ms^(-2)` along `y`-axisC. `1.2xx10^(-3) ms^(-2)` along `y` -axisD. `0.6xx10^(-3)ms^(-2)` along `y`- axis
Answer» Correct Answer - B

Discussion

227.	A thin non conducting disc of radius `R` is rotating clockwise (see figure) with an angular velocity `omega` about its central axis which is perpendicular to its plane Both its surfaces carry+ve charges of unifrom surface density Half the disc is in a region of a unifrom unidirectional magnetic field `B` parallel to the plane of the disc as shown Then .A. The net torque on the disc is zeroB. The net torque vector on the disc is directed leftwards .C. The net torque vector on the disc is directed rightwards .D. The net torque vector on the disc is parallel to `B` .
Answer» Correct Answer - B

Discussion

228.	A long cylidrical conductor of radius `R` carries a current i as shown in figure. The current desity `J` is a function of radius according to `J=br`, where `b` is a constant. Find an expression for the magnetic field `B` a. at a distasnce `r_1ltR` and b.at a distance `r_2gtR,` measured from the axis.
Answer» Correct Answer - `B_(1) = (mu_(0)br_(1)^(2))/(3),B_(2)=(mu_(0)(bR)^(3))/(3r_(2))` `ointoversetrarrB doversetrarrl = mu_(0) SigmaI` `B_(1) 2pi r_(1) = mu_(0) underset(0)overset(r_(1))intjdA =mu_(0) underset(0)overset(r_(1))intbr2pirdr` ` =mu_(0) 2pib(r^(3)/(3))_(0)^(r1) implies B_(1) = (mu_(0)r__1^(2)b)/(3)` `ointoversetrarrB doversetrarrl = mu_(0) SigmaI` `B 2pi r_(2) = mu_(0) underset(0)overset(R)intJdA =mu_(0) underset(0)overset(R)intbr2pirdr` `B_(2).2pir_(2) = mu_(0) 2pib((R^(3))/(3))implies B_(2) = (mu_(0)bR^(3))/(3r_(2))`

Discussion

229.	A proton carrying `1 MeV` kinetic energy is moving in a circular path of radius `R` in unifrom magentic field. What should be the energy of an `alpha-` particle to describe a circle of the same radius in the same field?A. `2MeV`B. `1MeV`C. `0*5MeV`D. `4MeV`
Answer» Here, `K.E._("proton")=1MeV`, Also `K.E.=1/2mv^2` So, `1/2mv^2=K.E`., So `mv^2=2K.E.implies mv=sqrt(2m(K.E.))` Then `(mv^2)/(R)=Bqv R=(mv)/(Bq)=(sqrt(2m(K.E.)))/(Bq)` So `K.E._P=(q^2pB^2R^2)/(2m_p)=(e^2B^2R^2)/(2m)` For `alphararr q_alpha=2e, m_alpha=4m` proton `K.E._alpha=(q_alpha^2B^2R^2)/(2m_alpha)=((2e)^2R^2B^2)/(2(4m))=(e^2R^2B^2)/(2m)` `(K.E._alpha)/(K.E._("proton"))=1` or `K.E._alpha=1xxK.E._("proton")` `=1xx1MeV=1MeV`

Discussion

230.	An electron (mass `=9.0xx10^(-31)` kg and charge `=1.6xx10^(-19)` coulomb) is moving in a circular orbit in a magnetic field of `1.0xx10^(-4)"weber"//m^(2)`. Its perido of revolution isA. `3.5xx10^(-7)sec`B. `7.0xx10^(-7)sec`C. `1.05xx10^(-6)sec`D. `2.1xx10^(-6)sec`
Answer» Correct Answer - A

Discussion

231.	An electron (mass `=9.1xx10^(-31)kg`, charge `=1.6xx10^(-19)C`) experiences no deflection if subjected to an electric field of `3.2x10^(5)V/m`, and a magnetic fields of `2.0xx10^(-3)Wb/m^(2)`. Both the fields are normal to the path of electron and to each other. If the electric field is removed, then the electron will revolve in an orbit of radiusA. `45m`B. `4.5m`C. `0.45m`D. `0.045m`
Answer» Correct Answer - C

Discussion

232.	An electron (mass `=9.1xx10^(-31)kg`, charge `=1.6xx10^(-19)C`) experiences no deflection if subjected to an electric field of `3.2x10^(5)V/m`, and a magnetic fields of `2.0xx10^(-3)Wb/m^(2)`. Both the fields are normal to the path of electron and to each other. If the electric field is removed, then the electron will revolve in an orbit of radiusA. `45m`B. `4.5m`C. `0.45m`D. 0.045m`
Answer» Correct Answer - C `oversetrarr(F_(e)) + oversetrarr(F_(m)) = 0 implies F_(e) = F_(m)` `eE = eVB` `v = (E)/(B) implies (3.2 xx 10^(5))/(2 xx 10^(-3)) = 1.6 xx 10^(8) m//s` `r =(mv)/(qB) = (9.1 xx 10^(-3) xx 1.6 xx 10^(8))/(1.6 xx 10^(-19) xx 2 xx 10^(-3)) = 0.45 m` .

Discussion

233.	A solenoid of length `50cm`, having `100` turns carries a current of `2*5A`. Find the magnetic field, (a) in the interior of the solenoid, (b) at one end of the solenoid. Given `mu_0=4pixx10^-7WbA^-1m^-1`.
Answer» Here, `I=25A`, `n=100//050=200m^-1` (a) `B=mu_0nI=4pixx10^-7xx200xx25` `=628xx10^-4T` (b) `B=(mu_0nI)/(2)=(4pixx10^-7xx200xx25)/(2)` `=314xx10^-4T`

Discussion

234.	A solenoid of length `020m`, having `120` turns carries a current of `25A`. Find the magnetic field: (a) in the interior of the solenoid, (b) at one end of the solenoid. Given `mu_0=4pixx10^-7` S.I. units.
Answer» Correct Answer - (a) `1885xx10^-3T` (b) `09425xx10^-3T` (i) `B=mu_0(NI)/(l)`, Here, `l=020m`, `N=120`, `I=25A` (ii) `B=(mu_0NI)/(2l)`

Discussion

235.	An `alpha`-particle is moving in a magnetic field of `(3hati+2hatj)` tesla with a velocity of `(5xx10^5hati)ms^-1`. What will be the magnetic force acting on the particle?
Answer» Here, `vecB=(3hati+2hatj)` and `vecv=(5xx10^5hati)`, `q=2e`. `vecF=q(vecvxxvecB)=q([5xx10^5hati)xx(3hati+2hatj)]=qxx10^6hatk=2exx10^6hatk=2xx16xx10^-19xx10^6hatk` or `\|vecF\|=32xx10^-13N`, towards positive z-direction.

Discussion

236.	When a certain length of wire is turned into one circular loop, the magnetic induction at the centre of coil due to some current flowing is `B_(1)` If the same wire is turned into three loops to make a circular coil, the magnetic induction at the center of this coil for the same current will beA. `B_(1)`B. `9B_(1)`C. `3B_(1)`D. `27B_(1)`
Answer» Correct Answer - B

Discussion

237.	`AB` and `CD` are long straight conductor, distance `d` apart, carrying a current `I`. The magnetic field at the midpoint of `BC` is A. `(-mu_(0)I)/(2pid)hatk`B. `(-mu_(0)I)/(pid)hatk`C. `(-mu_(0)I)/(4pid)hatk`D. `(-mu_(0)I)/(8pid)hatk`
Answer» Correct Answer - B

Discussion

238.	A current i is flowing in a straight conductor of length L. The magnetic induction at a point distant `L/4` from its centre will be-A. `(4mu_(0)i)/(sqrt(5)piL)`B. `(mu_(0)i)/(2piL)`C. `(mu_(0)i)/(sqrt(2)L)`D. Zero
Answer» Correct Answer - A

Discussion

239.	A charged particle is moving on a circular path of radius R in a uniform magnetic field under the Lorentz force F. How much work is done by the force in one round? Is the momentum of the particle changing?
Answer» When a charged particle moving on a circular path of radius R in a uniform magnetic field, the Lorentz magnetic force `F(=qvB)` acting on the particle, provides the required centripetal force for its circular motion. It means the Lerentz force acts along the radius towards the centre of circular path. While moving on a circular path, the small displacement `dvecr` of the charged particle is always perpendicular to Lerentz force, i.e., `theta=90^@`, therefore work done `dW=vecF.dvecr=Fdr cos 90^@=0`. Since the velocity of the charged particle, moving on a circular path is acting tangentially to the path whose direction is changing continuously in circular motion of the particle, therefore, the momentum of the particle is changing.

Discussion

240.	An electron `(q = 1.6 × 10^(-19) C)` is moving at right angles to a uniform magnetic field of `3.534 xx10^(-5)` T The time taken by the electron to complete a circular orbit isA. `2 mus`B. `4 mus`C. `3 mus`D. `1 mus`
Answer» Correct Answer - D

Discussion

241.	In case Hall effect for a strip having charge `Q` and area of crosssection `A`, the Lorentz force isA. Directly proportional to `Q`B. Inversely proportional to `Q`C. Inversely proportional to `A`D. Directly proportional to `A`
Answer» Correct Answer - A

Discussion

242.	What is meant by figure of merit of a galvanometer?
Answer» Figure of merit of galvanometer is defined as the amount of current which produces one scale deflection in the galvanometer.

Discussion

243.	What is the resistance of ideal ammeter and ideal voltmeter?
Answer» Correct Answer - Zero, infinite.

Discussion

244.	An iron bar magneti is heated to `1000^@C` and then cooled in a magnetic field free space. Will it retain magnetism?
Answer» No, it will not retain magnetism. Curie temperature for iron is about `750^@C`. At this temperature, it will become paramagnetic and lose its magnetism on heating further. Hence it will not retain magnetism on cooling in a mangetic field free space.

Discussion

245.	Two galvanometers A and B require 3mA and 5mA respectively to produce the same deflection of 10 division thenA. A is more sensitive than BB. B is more sensitive than AC. A and B are equally sensitiveD. Sensitivities of `B` is `5//3` times that of `A`
Answer» Correct Answer - A

Discussion

246.	Current flows through uniform square frames as shown. In which case is the magnetic field at the centre of the frame not zero ?A. B. C. D.
Answer» Correct Answer - C

Discussion

247.	Let the magnetic field on earth be modelled by that of a point magnetic dipole at the centre of earth. The angle of dip at a point on the geographical equatorA. is always zeroB. can be zero at specific pointsC. can be positive or negativeD. is bounded
Answer» Correct Answer - B::C::D The angle of dip at a point on the geographical equator satisfies the options (b), (c) and (d).

Discussion

248.	A magnetised needle of magnetic moment `4*8xx10^-2JT^-1` is placed at `30^@` with the direction of uniform magnetic field of magnitude `3xx10^-2T`. What is the torque acting on the needle?
Answer» Correct Answer - `72xx10^-4N-m` Here, `M=48xx10^-2JT^-1`, `theta=30^@`, `B=3xx10^-2T`, `tau=?` `tau=MBsin theta=48xx10^-2xx3xx10^-2xxsin30^@` `=72xx10^-4N-m`

Discussion

249.	Why is a current loop considered a magnetic dipole?
Answer» Like a bar magnet/magnetic dipole, a current loop possesses magnetic moment `M=NIA`. In an external magnetic field, the current loop experiences a torque and aligns its axis parallel to magnetic field.

Discussion

250.	Two infinitely long parallel wires carry equal current in same direction. The magnetic field at a mid point in between the two wires isA. Twice the magnetic field produced due to each of the wiresB. Half of the magnetic field produced due to each of the wiresC. Square of the magnetic field produced due to each of the wiresD. Zero
Answer» Correct Answer - D

Discussion

Explore topic-wise InterviewSolutions in .

A galvanometer with resistance `1000Omega` gives full scale deflection at `0*1mA`. What value of resistance should be added to `1000Omega` to increase its current range `10A`.A. `0*01Omega` in seriesB. `0*01Omega` in parallelC. `0*10Omega` in seriesD. `0*10Omega` in parallel

A galvanometer coil has a resistance of `12Omega` and the meter shows full scale deflection for a current of `3mA`. How will you convert the meter into a voltmeter of range 0 to 18V?

A galvanometer with a coil of resistance `12Omega` shows a full scale deflection for a current of `2*5mA`. How will you convert it into a voltmeter of range `7*5V`? Also, find the total resistance of voltmeter formed.

A galvanometer coil has a resistance of `15Omega` and the meter shows full scale deflection for a current of `4mA`. How will you convert the meter into an ammeter of range 0 to 6A?

In a hydrogen atom, the electron moves in an orbit of radius `0*5 Å`, making `10^(16)rps`. Calculate the magnetic moment associated with the orbital motion of electron.

A long wire bent as shown in fig. carries current I. If the radius of the semicircular portion is a, the magnetic field at centre C is A. `(mu_0I)/(4r)`B. `(mu_0I)/(4pir)sqrt(pi^2+4)`C. `(mu_0I)/(4pi)+(mu_0I)/(2pir)`D. `(mu_0I)/(4pir)sqrt(pi^2-4)`

A stream of protons is moving parallel to a stream of electrons. Will the two stream tend to come closer or move apart?

Two long parallel wires are at a distance ` 2d` apart. They carry steady equal currents flowing out of the plane of the paper , as shown. The variation of the magnetic field `B` along the line `XX` is given byA. B. C. D.

Two wires are held perpendicular to the plane of paper and are 5 m apart. They carry currents of 2.5 A and 5 A in same direction. Then, the magnetic field strength (B) at a point midway between the wires will beA. `(mu_(0))/(4pi)T`B. `(mu_(0))/(2pi)T`C. `(3mu_(0))/(2pi)T`D. `(3mu_(0))/(4pi)T`

Two straight parallel wires, both carrying `10` ampere in the same direction attract each other with a force of `1xx10^(-3)N`. If both currents are doubled, the force of attraction will beA. `1xx10^(-3)N`B. `2xx10^(-3)N`C. `4xx10^(-3)N`D. `0.25xx10^(-3)N`

Three long straight and parallel wires carrying currents are arranged as shown in the figure. The wire `C` which carreis a current of `5.0 amp` is so placed that it experiences no force. The distance of wire `C` from `D` is then A. `9cm`B. `7cm`C. `5cm`D. `3cm`

A circular loop has a radius of `5 cm` and it is carrying a current of `0.1 amp`. It magneitc moment isA. `1.32xx10^(-4)amp-m^(2)`B. `2.62xx10^(-4)amp-m^(2)`C. `5.25xx10^(-4)amp-m^(2)`D. `7.85xx10^(-4)amp-m^(2)`

A long straight wire carries a current of `pi amp`. The magnetic field due to it will be `5xx10^(-5)"weber"//m^(2)` at what distance from the wire [`mu_(0)=` permeability of air]A. `10^(4)mu_(0)` metreB. `(10^(4))/(mu_(0))` metreC. `10^(6)mu_(0)` metreD. `(10^(6))/(mu_(0))` metre

The two linear parallel conductors carrying currents in the opposite direction…………………each other.

A straight wire carrying a cureent `i_(1)` amp runs along the axis of a circular current `i_(2)` amp. Then the force of interaction between the two current carrying conductors isA. `oo`B. Zero at the centre of loopC. `(mu_(0))/(4pi)(2i_(1)i_(2))/rN//m`D. `(2i_(1)i_(2))/r N//m`

An arbitrary shaped closed coil is made of a wire of length `L` and a current `I` ampere is flowing in it. If the plane of the coil is perpendicular to megnetic field `vecB`, the force on the coil isA. ZeroB. `lBL`C. `2lBL`D. `1/2 IBL`

A unifrom conducting wire `ABC` has a mass of `10g`. A current of `2 A` flows through it. The wire is kept in a unifrom magnetic field `B = 2T`. The accleration of the wire will be A. ZeroB. `12 ms^(-2)` along `y`-axisC. `1.2xx10^(-3) ms^(-2)` along `y` -axisD. `0.6xx10^(-3)ms^(-2)` along `y`- axis

A proton carrying `1 MeV` kinetic energy is moving in a circular path of radius `R` in unifrom magentic field. What should be the energy of an `alpha-` particle to describe a circle of the same radius in the same field?A. `2MeV`B. `1MeV`C. `0*5MeV`D. `4MeV`

An electron (mass `=9.0xx10^(-31)` kg and charge `=1.6xx10^(-19)` coulomb) is moving in a circular orbit in a magnetic field of `1.0xx10^(-4)"weber"//m^(2)`. Its perido of revolution isA. `3.5xx10^(-7)sec`B. `7.0xx10^(-7)sec`C. `1.05xx10^(-6)sec`D. `2.1xx10^(-6)sec`

A solenoid of length `50cm`, having `100` turns carries a current of `2*5A`. Find the magnetic field, (a) in the interior of the solenoid, (b) at one end of the solenoid. Given `mu_0=4pixx10^-7WbA^-1m^-1`.

A solenoid of length `0*20m`, having `120` turns carries a current of `2*5A`. Find the magnetic field: (a) in the interior of the solenoid, (b) at one end of the solenoid. Given `mu_0=4pixx10^-7` S.I. units.

An `alpha`-particle is moving in a magnetic field of `(3hati+2hatj)` tesla with a velocity of `(5xx10^5hati)ms^-1`. What will be the magnetic force acting on the particle?

`AB` and `CD` are long straight conductor, distance `d` apart, carrying a current `I`. The magnetic field at the midpoint of `BC` is A. `(-mu_(0)I)/(2pid)hatk`B. `(-mu_(0)I)/(pid)hatk`C. `(-mu_(0)I)/(4pid)hatk`D. `(-mu_(0)I)/(8pid)hatk`

A current i is flowing in a straight conductor of length L. The magnetic induction at a point distant `L/4` from its centre will be-A. `(4mu_(0)i)/(sqrt(5)piL)`B. `(mu_(0)i)/(2piL)`C. `(mu_(0)i)/(sqrt(2)L)`D. Zero

A charged particle is moving on a circular path of radius R in a uniform magnetic field under the Lorentz force F. How much work is done by the force in one round? Is the momentum of the particle changing?

An electron `(q = 1.6 × 10^(-19) C)` is moving at right angles to a uniform magnetic field of `3.534 xx10^(-5)` T The time taken by the electron to complete a circular orbit isA. `2 mus`B. `4 mus`C. `3 mus`D. `1 mus`

In case Hall effect for a strip having charge `Q` and area of crosssection `A`, the Lorentz force isA. Directly proportional to `Q`B. Inversely proportional to `Q`C. Inversely proportional to `A`D. Directly proportional to `A`

What is meant by figure of merit of a galvanometer?

What is the resistance of ideal ammeter and ideal voltmeter?

An iron bar magneti is heated to `1000^@C` and then cooled in a magnetic field free space. Will it retain magnetism?

Two galvanometers A and B require 3mA and 5mA respectively to produce the same deflection of 10 division thenA. A is more sensitive than BB. B is more sensitive than AC. A and B are equally sensitiveD. Sensitivities of `B` is `5//3` times that of `A`

Current flows through uniform square frames as shown. In which case is the magnetic field at the centre of the frame not zero ?A. B. C. D.

Let the magnetic field on earth be modelled by that of a point magnetic dipole at the centre of earth. The angle of dip at a point on the geographical equatorA. is always zeroB. can be zero at specific pointsC. can be positive or negativeD. is bounded

A magnetised needle of magnetic moment `4*8xx10^-2JT^-1` is placed at `30^@` with the direction of uniform magnetic field of magnitude `3xx10^-2T`. What is the torque acting on the needle?

Why is a current loop considered a magnetic dipole?

A galvanometer with resistance `1000Omega` gives full scale deflection at `01mA`. What value of resistance should be added to `1000Omega` to increase its current range `10A`.A. `001Omega` in seriesB. `001Omega` in parallelC. `010Omega` in seriesD. `0*10Omega` in parallel

A galvanometer with a coil of resistance `12Omega` shows a full scale deflection for a current of `25mA`. How will you convert it into a voltmeter of range `75V`? Also, find the total resistance of voltmeter formed.

A solenoid of length `020m`, having `120` turns carries a current of `25A`. Find the magnetic field: (a) in the interior of the solenoid, (b) at one end of the solenoid. Given `mu_0=4pixx10^-7` S.I. units.