624 + Interview Questions in MAGNETIC-EFFECT-OF-CURRENT IN GENERAL AWARENESS Page 1 InterviewSolution

1.	A current carrying loop consists of 3 identical quarter circles of radius R, lying in the positive quadrants of the x-y, y-z and z-x planes with their centres at the origin, joined together. Find the direction and magnitude to `vecB` at the origin.
Answer» `vecB=vecB_(xy)+vecB_(yz)+vecB_(zx)=((mu_0)/(4pi)(I)/(R)pi/2hatk)+((mu_0)/(4pi)I/Rpi/2hati)+((mu_0)/(4pi)I/Rpi/2hatj)=(mu_0)/(4pi)I/Rpi/2(hati+hatj+hatk)` `=(mu_0I)/(8R)(hati+hatj+hatk)`

Discussion

2.	A long straight wire carries a current of `2A`. An electron travels with a velocity of `4xx10^4ms^-1` parallel to the wire `01mm` from it, and in a direction opposite to the current. What force does the magnetic field of current exert on the moving electron. Charge on electron `=16xx10^(-19)C`.
Answer» Correct Answer - `256xx10^(-20)N`, towards the conductor `B=(mu_0)/(4pi)(2I)/(r)=10^-7xx(2xx2)/(01)=4xx10^-6T` When electron travles parallel to the conductor, `vecB` is perpendicular to `vecv`(i.e. `theta=90^@`) `:. F=evBsintheta` `=16xx10^(-19)xx4xx10^4xx4xx10^-6xxsin90^@` `=256xx10^(-20)N` This force will be acting towards the conductor.

Discussion

3.	The variation of magnetic susceptibility `chi` with the temperature T of a ferromagnetic material can be plotted asA. B. C. D.
Answer» Correct Answer - B The magnetic susceptibility `(chi)` of a ferromagnetic substance decrease with increase in temperature i.e. `chiprop1//T`. Hence, option (b) is true.

Discussion

4.	A wire of length L metre carrying a current of I ampere is bent in the form of a circle. Find its magnetic moment.
Answer» `L=2piR` or `R=L//2pi`, Magnetic moment `=IA=IpiR^2` `=Ipi((L)/(2pi))^2=IL^2//4pi`

Discussion

5.	A wire of length `L metre ` , carrying a current `I `ampere is bent in the form of a circle . The magnitude of its magnetic moment is …………. ` MKS units ` .A. `(IL)/(4pi)`B. `(IL^(2))/(4pi)`C. `(I^(2)L^(2))/(4pi)`D. `(I^(2)L)/(4pi)`
Answer» Correct Answer - B

Discussion

6.	A wire of length `L metre ` , carrying a current `I `ampere is bent in the form of a circle . The magnitude of its magnetic moment is …………. ` MKS units ` .
Answer» If r is radius of the circle, then `2pir=L` or `r=(L)/(2pi)` `M=IA=Ipir^2=Ipi((L)/(2pi))^2=(IL^2)/(4pi)`

Discussion

7.	A long wire carries a steady curent . It is bent into a circle of one turn and the magnetic field at the centre of the coil is `B`. It is then bent into a circular loop of `n` turns. The magnetic field at the centre of the coil will beA. nBB. `n^(2)B`C. 2nBD. `2n^(2)B`
Answer» Correct Answer - b

Discussion

8.	A long solenoid has a radius `a` and number of turns per unit length is `n`. If it carries a current i, then the magnetic field on its axis is directly proportional toA. `ani`B. `ni`C. `(ni)/a`D. `n^(2)i`
Answer» Correct Answer - B

Discussion

9.	Two concentric circular coils of ten turns each are situated in the same plane. Their radii are `20` and `40 cm` and they carry respectively `0.2` and `0.3` ampere current in opposite direction. The magnetic field in `Wb//m^(3)` at the centre isA. `35/4 mu_(0)`B. `(mu_(0))/80`C. `7/80 mu_(0)`D. `5/4 mu_(0)`
Answer» Correct Answer - D

Discussion

10.	A particle of mass `m` and charge `q` moves with a constant velocity `v` along the positive `x` direction. It enters a region containing a uniform magnetic field `B` directed along the negative `z` direction, extending from `x = a` to `x = b`. The minimum value of `v` required so that the particle can just enter the region `x gt b` isA. `qb B//m`B. `q(b-a)B//m`C. `qa B//m`D. `a(b+a)B//2m`
Answer» Correct Answer - B

Discussion

11.	Two particles `A and B` of masses ` m_(A) and m_(B)` respectively and having the same charge are moving in a plane . The speeds of the particles are ` v_(A) and v_(B)` respectively and the trajectories are as shown in the figure. Then A. `m_(A)v_(A)ltm_(B)v_(B)`B. `m_(A)v_(A)gtm_(B)v_(B)`C. `m_(A)ltm_(B)` and `v_(A)lt v_(B)`D. `m_(A)=m_(B)` and `v_(A)=v_(B)`
Answer» Correct Answer - B

Discussion

12.	Two ions having masses in the ratio `1 : 1` and charges `1 : 2` are projected into uniform magnetic field perpendicular to the field with speeds in th ratio `2 : 3`. The ratio of the radius of circular paths along which the two particles move isA. `4:3`B. `2:3`C. `3:1`D. `1:4`
Answer» Correct Answer - A

Discussion

13.	Two particles `A and B` of masses ` m_(A) and m_(B)` respectively and having the same charge are moving in a plane . The speeds of the particles are ` v_(A) and v_(B)` respectively and the trajectories are as shown in the figure. Then A. `m_(A)V_(A) ltm_(B)V_(B)`B. `m_(A)V_(A) gt m_(B)V_(B)`C. `m_(A) lt m_(B) ` and `V_(A) lt V_(B)`D. `m_(A) = m_(B) ` and `V_(A) = V_(B)`
Answer» Correct Answer - `B` `r = (mv)/(qB)` Givent that `r_(A) gt r_(B)` `m_(A) V_(A) gtm_(V) V_(B)` .

Discussion

14.	A charged particle with charge `q` enters a region of constant, uniform and mututally orthogonal fields `vec(E) and vec(B)` with a velocity `vec(v)` perpendicular to both `vec(E) and vec(B)`, and comes out without any change in magnitude or direction of `vec(v)`. ThenA. `vecv=vecBxxvecE//E^2`B. `vecv=vecExxvecB//B^2`C. `vecv=vecBxxvecE//B^2`D. `vecv=vecExxvecB//E^2`
Answer» Correct Answer - B When charged particle goes undefected, then `qE=qvB` or `v=E//B` The two forces acting on moving charged particle due to electric and magnetic fields oppose each other if `vecv` is along `vecExxvecB` Now, `v=(EB)/(B^2)=(EB sin 90^@)/(B^2)=(\|vecExxvecB\|)/(B^2)` `:. vecv=((vecExxvecB))/(B^2)`

Discussion

15.	A beam of electrons is moving with constant velocity in a region having simultaneous perpendicular electric and magnetic fields of strength `20Vm^(-1)` and 0.5 T, respectively at right angles to the direction of motion of the electrons. Then, the velocity of electrons must beA. `8 m//s`B. `20 m//s`C. `40 m//s`D. `(1)/(40)m//s`
Answer» Correct Answer - c

Discussion

16.	A non - popular loop of conducting wire carrying a current `I` is placed as shown in the figure . Each of the straighrt sections of the loop is of the length ` 2a`. The magnetic field due to this loop at the point `P(a, 0 ,a)` points in the direction A. `1/(sqrt(2))(-hatj+hatk)`B. `1/(sqrt(3))(-hatj+hatk+hati)`C. `1/(sqrt(3))(hati+hatj+hatk)`D. `1/(sqrt(2))(hati+hatk)`
Answer» Correct Answer - D

Discussion

17.	Two short magnets (1) and (2) of magnetic moments `2Am^2` and `5Am^2` respectively are placed along two lines drawn at `90^@` to each other as shown in figure. At the point of intersection of their axes, the magnitude of magnetic field is `nxx10^-5T`. What is n?
Answer» Correct Answer - `(2)` Here, `M_1=2Am^2`, `d_1=03m` `M_2=5Am^2`, `d_2=04m` The point of intersection P lies on axial line of the two magnets, figure. `:. B_1=(mu_0)/(4pi)xx(2M_1)/(d_1^3)=10^-7xx(2xx2)/((03)^3)` `=148xx10^-5T` along `S_1N_1` `B_2=(mu_0)/(4pi)xx(2M_2)/(d_2^3)=10^-7xx(2xx5)/((04)^3)` `=156xx10^-5T` along `S_2N_2` As `B_1` and `B_2` are perpendicular to each other, therefore resultant magnetic field at P is `B=sqrt(B_1^2+B_2^2)` `=sqrt((148xx10^-5)^2+(156xx10^-5)^2)` `=215xx10^-5T=nxx10^-5T` where `n=215`, which can be taken as 2

Discussion

18.	Two short magnets having magnetic moments in the ratio `8:1`, when placed on opposite sides of a deflection magnetometer produce no deflection. If distance of stronger magnet from the centre of deflection magnetometer is `6cm`, what is the distance of weaker magnet from the centre?
Answer» Here, `M_1/M_2=8/1, d_1=6cm d_2=?` For short magnets, In the null method `M_1/M_2(d_1/d_2)^3=8/1=(2)^3` `:. d_1/d_2=2, d_2=d_1/2=6/2=3cm`

Discussion

19.	The hyterisis loss for a sample of `12kg` is `300Jm^-3cycl e^-1`. If density of iron is `7500kg//m^3`, calculate energy loss per hour at `50cycl e//sec`.
Answer» Here, energy loss/volume, per cycle `=300Jm^-3cycl e^-1`. Volume of sample `=(mass)/(density)=(12)/(7500)m^3` Total energy loss/cycle`=300xx(12)/(7500)=12/25` Energy loss/ sec `=12/25xx50=24J//sec` Energy loss/hour`=24xx60xx60J=8*64xx10^4J`

Discussion

20.	The hysterisis loss of a sample of iron is `300Jm^-3 cycl e^-1`. If density of iron is `7500kg//m^3` and mass of iron piece is `15kg`, calculate energy lost/hour when the frequency of magnetisation/demagnetisation used in `50c//s`.
Answer» Correct Answer - `54xx10^4J//hour` Here, energy loss due to hysterisis `=300Jm^-3cycl e^-1` `rho=7500kg//m^3, m=15kg, v=50c//s` Energy lost/hour=? `V=m/rho=(15)/(7500)m^3=2xx10^-3m^3` Energy lost/cycle =`300xx2xx10^-3=06` Energy lost/sec`=06xx50=30J//s` Energy lost/hour=`30xx60xx60` `=54xx10^4J//hr`

Discussion

21.	Hysterisis loss for a sample of `120kg` is `300Jm^-3//cycl e`. Frequency of supply is `50cycl e//s`. If density of iron is `7500kg//m^3`, find energy loss in 60 minutes.
Answer» Here, `m=120kg` Hysterisis loss `=300Jm^-3//cycl e`. Frequency `=50c//s` Density of iron `=7500kg//m^3` Total time =`60 mins=60xx60s` Volume of specimen `=(mass)/(density)=(120)/(7500)m^3` Loss of Loss of `en ergy//cycl e=300xx(120)/(7500)=360/75` Loss of `en ergy//sec=360/75xx50` Total loss of energy `=360/75xx50xx60xx60` `=240xx3600` `=86*4xx10^4J`

Discussion

22.	Ferro magnetic materials used in transformer must haveA. low permeability and high hysteresis lossB. high permeability and low hysteresis lossC. high permeability and high hysteresis lossD. low permeability and low hysteresis loss
Answer» Correct Answer - B A ferromagnetic materials core used in transfarmer must have high permeability so that it may show high value of magnetic induction for the given magnetising field. It should have low hysteresis loss in order to decrease the energy losses.

Discussion

23.	Assertion: Cyclotron is a device which is used to accelerate the position ions. Reason: Cyclotron frequency depends upon the velocity.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 - C

Discussion

24.	Cyclotron is used to accelerateA. ElectronsB. NeutronsC. Positive ionsD. Negative ions
Answer» Correct Answer - C

Discussion

25.	In a cyclotron, a charged particleA. undergoes acceleration all the time.B. speeds up between the dees because of the magnetic fieldC. speeds up in a deeD. slows down with a dee and speeds up between dees
Answer» Correct Answer - A In a cyclotron, a charged particle describes the circular paths inside the dees and is accelerated while going from one dee to another due to electric field. During circular path, the charged particle has centripetal acceleration which is provided by the magnetic force due to magnetic field. Therefore, a charged particle undergoes acceleration all the time, inside the cyclotron.

Discussion

26.	A cyclotron is not suitable to accelerate electrons. Why?
Answer» When an electron is accelerated in a cyclotron, very soon it acquries a very high velocity. Due to it, its mass increases with velocity according to relation `m=m_0//sqrt((1-v^2//c^2))`. Therefore the time taken by electron to describe semicircular path inside the dee of a cyclotron, `t=pim//Bq`, also increases with the increases of m. Due to it, the electron does not arrive in the gap between the two dees exactly at the instant, the polarity of the two dees is reversed. As a result of it, the electron goes out of steps with the oscillating electric field and hence can not be accelerated by cyclotron.

Discussion

27.	In cyclotron, an ion is made to travel successively along semicircles of increasing radius under the action of a magnetic field. The angular velocity of the ion is independent ofA. speed of ionB. radius of the circleC. mass of the ionD. charge of the ion
Answer» Correct Answer - B Here, as wer know is cyclotron centripetal force on charge is equal to magnetic force `F_c=F_m` So `(mv^2)/(r)=qvBimpliesv/r=(qB)/(m)` [As `v=romega`, So `V/r=omega`] So `omega=(qB)/(m)` [i.e., Independent of v, r]

Discussion

28.	Verify that the cyclotron frequency `omega=eB//m` has the correct dimensions of `[T]^-1`.
Answer» If a charged particle is moving perpendicular to the magnetic field, then `qvB=(mv^2)/(r)` or `(qB)/(m)=(v)/(r)=omega` `:.` Dimensions of `omega=(qB)/(m)=v/r=(LT^-1)/(L)=[T^-1]`

Discussion

29.	What is the basic difference between the atom/molecule of a diamagnetic and paramagnetic material?
Answer» The atom/molecules of a diamagnetic substance do not posses any net magnetic moment on their own. In paramagnetic substances, each individual atom/molecule has non zero magnetic moment of its own.

Discussion

30.	In what way is the behaviour of a diamagnetic material different from that of a paramagnetic material, when kept in an external magnetic field?
Answer» In a diamagnetic material, the magnetic field lines of external field avoid to enter, whereas in a paramagnetic material, the magnetic field lines tend to pass through it.

Discussion

31.	Identify the materials, which can be classified as paramagnetic and diamagnetic: Al, Bi, Cu, Na.
Answer» Aluminium and sodium are paramagnetic Bismuth and copper are diamagnetic.

Discussion

32.	The magnetic permeability of a substance is `7*5xx10^-3TmA^-1`. Find relative permeability and magnetic susceptibility of the substance.
Answer» Here, `mu=75xx10^-3 mu_r=? chi_m=?` `mu_r=(mu)/(mu_0)=(75xx10^-3)/(4pixx10^-7)=06xx10^4` From `mu_r=1+chi_m` `chi_m=mu_r-1` `chi_m=06xx10^4-1=5999`

Discussion

33.	Which of the following substances are paramagnetic? Al, Bi, Cu, Ca, Pb, Ni.
Answer» Aluminium and calcium are paramagnetic.

Discussion

34.	Explain quantitatively the order of magnitude difference between the diamagnetic susceptibility of `N_2(-5xx10^-9)` (at STP) and `Cu(-10^-5)`.
Answer» Density of nitrogen, `rho_(N_2)=(28g)/(22.4litre)=(28g)/(22400c c)` We know, density of copper, `rho_(cu)=8g//c c` `:. (rho_(N_2))/(rho_(Cu))=(28)/(22400)xx1/8=16xx10^-4`. Given, `(chi_(N_2))/(chi_(cu))=(5xx10^-9)/(10^-5)=5xx10^-4` …(i) We know that `chi= ("intensity of magnetisation"(I))/("magnetising force" (H))=I/H=("magnetic moment"(M)//volume(V))/(H)` `=(M)/(HV)=(M)/(H["mass"(m)//"density"(rho)])=(Mrho)/(Hm)` `:. chiproprho` for the given value of `M//Hm`. Hence, `(chi_(N_2))/(chi_(Cu))=(rho_(N_2))/(rho_(Cu))=16xx10^-4` ...(ii) From (i) and (ii), we note that the major difference in susceptibility of `N_2` and `Cu` is due to their density.

Discussion

35.	Assertion: A current `I` flows I flows along the length of an infinitely long straght and thin walled pipe. Then the magnetic field at any point inside the pipe is zero. Reason: `oint vec(B).d vec(l) =mu I`A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explaination of Statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true.
Answer» Correct Answer - A Here, both statement-1 and statement-2 are correct and statement-2 is the correct explanation of statement-1.

Discussion

36.	Satement -1: A beam of electron passes undeflected through region `oversetrarr(E)& oversetrarr(B)` In the region `oversetrarr(E)& oversetrarr(B)` are present and perpendicular to each other and the particle is moving perpendicular to both of them . Statement-2: In the region `vecE & vecB`, both are present and perpendicular to each other and the particle is moving perpendicular to both of them.
Answer» Correct Answer - `A`

Discussion

37.	An electron enters a region where electrostatic field is `20N//C` and magnetic field is `5T`. If electron passes undeflected through the region, then velocity of electron will beA. `0.25ms^(-1)`B. `2ms^(-1)`C. `4ms^(-1)`D. `8ms^(-1)`
Answer» Correct Answer - C

Discussion

38.	A proton moving withh a velocity `2.5xx10^(7)m//s` enters a magnetic field of intensity `2.5T` making an angle `30^(@)` with the magnetic field. The force on the proton isA. `3xx10^(-12)N`B. `5xx10^(-12)N`C. `6xx10^(-12)N`D. `9xx10^(-12)N`
Answer» Correct Answer - B

Discussion

39.	Maximum kinetic energy of the positive ion in the cyclotron isA. `(q^(2)Br_(0))/(2m)`B. `(qB^(2)R_(0))/(2m)`C. `(q^(2)B^(2)r_(0)^(2))/(2m)`D. `(qBr_(0))/(2m^(2))`
Answer» Correct Answer - C

Discussion

40.	The cyclotron frequency of an electron gyrating in a magnetic field of `1T` is approximately:A. `28 MHz`B. `280 MHz`C. `2.8GHz`D. `28GHz`
Answer» Correct Answer - D

Discussion

41.	Assertion: Free electron always keep on moving in a conductor even then no magnetic force act on them in magnetic field unless a current is passed through it. Reason: The average velocity of free electron is zero.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 - A

Discussion

42.	Assertion: The ion cannot move with a speed beyond a certain limit in a cyclotron. Reason: As velocity increases time taken by ion increases.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 - C

Discussion

43.	Two tangent galvanometers A and B have coils of radii `8cm` and `16cm` respectively and resistance `8ohm` each. They are connected in parallel to a cell of emf `4V` and negligible internal resistance. The deflections produced are `30^@` and `60^@` respectivley. A has 2 turns. What is the number of turns in B?
Answer» Here, `r_1=8cm, r_2=16cm`, `R_1=R_2=8ohm` `E=4V, theta_1=30^@, theta_2=60^@, n_1=2, n_2=?` `I=E/R=(2r_1H)/(mu_0n_1)tantheta_1=(2r_2H)/(mu_0n_2)=tan_2` `:. n_2=(n_1r_2tantheta_2)/(r_1tantheta_1)=(2xx16tan60^@)/(8tan30^@)=(4sqrt3)/(1//sqrt3)` `=12`

Discussion

44.	The coercitivity of a small magnet where the ferromagnet gets demagnetized is `3 xx 10^(3) Am^(-1)`. The current required to be passed in a solenoid of length `10 cm` and number of turns `100`, so that the magnet gets demagnetized when inside the solenoid , is :A. `3A`B. `6A`C. `30mA`D. `60mA`
Answer» Correct Answer - A Here, coercivity of given magnet `=B/mu_0=3xx10^3Am^-1` As `B=(mu_0NI)/(L)` or `I=(BL)/(mu_0N)=((3xx10^3)xx0*1)/(100)=3A`

Discussion

45.	The coercivity of a certain permanent magnet is `4*0xx10^4Am^-1`. This magnet is placed inside a solenoid `15cm` long and having `600turns`. Calculate the current required to demagnetise the magnet completely.
Answer» Here, coercivity `=40xx10^4Am^-1` of magnet implies that magnetising intensity `H=40xx10^4Am^-1` is required to be applied in opposite direction to demagnetise the magnet. n=number of turns per unit length `=(600)/(15xx10^-2)=4000turns//m` As `H=nI :. I=H/n=(4*0xx10^4)/(4000)=10A`

Discussion

46.	The magnetic field at a point near the centre but outside a current carrying solenoid is zero. Explain why?
Answer» When current is passed through a long straight solenoid, then each turn of the solenoid can be regarded as a circular loop carrying current and thus will be producting a magnetic field. The total magnetic field at a point is the vector sum of the magnetic fields due to currents through all the turns in the solenoid. At a point near the centre outside the solenoid, the magnetic fields due to neighbouring loops are equal in magnitude and opposite in direction. Hence the resultant magnetic field at a point near the centre outside the solenoid is zero.

Discussion

47.	What is the magnitude of force on a wire of length `004m` placed inside a solenoid near its centre, making an angle of `30^@` with its axis? The wire carries a current of `12A` and the magnetic field due to the solenoid is of magnitude `025T`.
Answer» Correct Answer - `006N` Here, `l=004m`, `theta=30^@`, `I=12A`, `B=025T`. Force, `F=IBlsin theta` `=12xx(025)xx(004)sin 30^@` `=12xx025xx004xx1//2=006N`

Discussion

48.	What is the potential energy of a dipole when it is perpendicular to a magnetic field?
Answer» `P.E.=-MB cos theta=-MB cos 90^@=zero`.

Discussion

49.	What does the torque do?
Answer» The torque tries to align the magnet along the direction of magnetic field `(thetararrZero)`.

Discussion

50.	Two identical magnetic dipoles of magnetic moments `1*0Am^2` each are placed at a separation of `2m` with their axes perpendicular to each other. What is the resultant magnetic field at a point midway between the dipoles?A. `5xx10^-7T`B. `sqrt5xx10^-7T`C. `10^-7T`D. `2xx10^-7T`
Answer» Correct Answer - B As the axes are perpendicular, mid point lies on axial line of one magnet and on equatorial line of other magnet. `:. B_1=(mu_0)/(4pi)(2M)/(d^3)=(10^-7xx2xx1)/(1^3)=2xx10^-7T` and `B_2=(mu_0)/(4pi)(M)/(d^3)=(10^-7xx1)/(1^3)=10^-7T`. As `B_1_\|_B_2`, `:.` Resultant field `=sqrt(B_1^2+B_2^2)=sqrt5xx10^-7T`

Discussion

Explore topic-wise InterviewSolutions in .

A current carrying loop consists of 3 identical quarter circles of radius R, lying in the positive quadrants of the x-y, y-z and z-x planes with their centres at the origin, joined together. Find the direction and magnitude to `vecB` at the origin.

A long straight wire carries a current of `2A`. An electron travels with a velocity of `4xx10^4ms^-1` parallel to the wire `0*1mm` from it, and in a direction opposite to the current. What force does the magnetic field of current exert on the moving electron. Charge on electron `=1*6xx10^(-19)C`.

The variation of magnetic susceptibility `chi` with the temperature T of a ferromagnetic material can be plotted asA. B. C. D.

A wire of length L metre carrying a current of I ampere is bent in the form of a circle. Find its magnetic moment.

A wire of length `L metre ` , carrying a current `I `ampere is bent in the form of a circle . The magnitude of its magnetic moment is …………. ` MKS units ` .A. `(IL)/(4pi)`B. `(IL^(2))/(4pi)`C. `(I^(2)L^(2))/(4pi)`D. `(I^(2)L)/(4pi)`

A wire of length `L metre ` , carrying a current `I `ampere is bent in the form of a circle . The magnitude of its magnetic moment is …………. ` MKS units ` .

A long wire carries a steady curent . It is bent into a circle of one turn and the magnetic field at the centre of the coil is `B`. It is then bent into a circular loop of `n` turns. The magnetic field at the centre of the coil will beA. nBB. `n^(2)B`C. 2nBD. `2n^(2)B`

A long solenoid has a radius `a` and number of turns per unit length is `n`. If it carries a current i, then the magnetic field on its axis is directly proportional toA. `ani`B. `ni`C. `(ni)/a`D. `n^(2)i`

Two ions having masses in the ratio `1 : 1` and charges `1 : 2` are projected into uniform magnetic field perpendicular to the field with speeds in th ratio `2 : 3`. The ratio of the radius of circular paths along which the two particles move isA. `4:3`B. `2:3`C. `3:1`D. `1:4`

Two short magnets (1) and (2) of magnetic moments `2Am^2` and `5Am^2` respectively are placed along two lines drawn at `90^@` to each other as shown in figure. At the point of intersection of their axes, the magnitude of magnetic field is `nxx10^-5T`. What is n?

Two short magnets having magnetic moments in the ratio `8:1`, when placed on opposite sides of a deflection magnetometer produce no deflection. If distance of stronger magnet from the centre of deflection magnetometer is `6cm`, what is the distance of weaker magnet from the centre?

The hyterisis loss for a sample of `12kg` is `300Jm^-3cycl e^-1`. If density of iron is `7500kg//m^3`, calculate energy loss per hour at `50cycl e//sec`.

The hysterisis loss of a sample of iron is `300Jm^-3 cycl e^-1`. If density of iron is `7500kg//m^3` and mass of iron piece is `15kg`, calculate energy lost/hour when the frequency of magnetisation/demagnetisation used in `50c//s`.

Hysterisis loss for a sample of `120kg` is `300Jm^-3//cycl e`. Frequency of supply is `50cycl e//s`. If density of iron is `7500kg//m^3`, find energy loss in 60 minutes.

Ferro magnetic materials used in transformer must haveA. low permeability and high hysteresis lossB. high permeability and low hysteresis lossC. high permeability and high hysteresis lossD. low permeability and low hysteresis loss

Cyclotron is used to accelerateA. ElectronsB. NeutronsC. Positive ionsD. Negative ions

In a cyclotron, a charged particleA. undergoes acceleration all the time.B. speeds up between the dees because of the magnetic fieldC. speeds up in a deeD. slows down with a dee and speeds up between dees

A cyclotron is not suitable to accelerate electrons. Why?

In cyclotron, an ion is made to travel successively along semicircles of increasing radius under the action of a magnetic field. The angular velocity of the ion is independent ofA. speed of ionB. radius of the circleC. mass of the ionD. charge of the ion

Verify that the cyclotron frequency `omega=eB//m` has the correct dimensions of `[T]^-1`.

What is the basic difference between the atom/molecule of a diamagnetic and paramagnetic material?

In what way is the behaviour of a diamagnetic material different from that of a paramagnetic material, when kept in an external magnetic field?

Identify the materials, which can be classified as paramagnetic and diamagnetic: Al, Bi, Cu, Na.

The magnetic permeability of a substance is `7*5xx10^-3TmA^-1`. Find relative permeability and magnetic susceptibility of the substance.

Which of the following substances are paramagnetic? Al, Bi, Cu, Ca, Pb, Ni.

Explain quantitatively the order of magnitude difference between the diamagnetic susceptibility of `N_2(-5xx10^-9)` (at STP) and `Cu(-10^-5)`.

An electron enters a region where electrostatic field is `20N//C` and magnetic field is `5T`. If electron passes undeflected through the region, then velocity of electron will beA. `0.25ms^(-1)`B. `2ms^(-1)`C. `4ms^(-1)`D. `8ms^(-1)`

A proton moving withh a velocity `2.5xx10^(7)m//s` enters a magnetic field of intensity `2.5T` making an angle `30^(@)` with the magnetic field. The force on the proton isA. `3xx10^(-12)N`B. `5xx10^(-12)N`C. `6xx10^(-12)N`D. `9xx10^(-12)N`

Maximum kinetic energy of the positive ion in the cyclotron isA. `(q^(2)Br_(0))/(2m)`B. `(qB^(2)R_(0))/(2m)`C. `(q^(2)B^(2)r_(0)^(2))/(2m)`D. `(qBr_(0))/(2m^(2))`

The cyclotron frequency of an electron gyrating in a magnetic field of `1T` is approximately:A. `28 MHz`B. `280 MHz`C. `2.8GHz`D. `28GHz`

The coercitivity of a small magnet where the ferromagnet gets demagnetized is `3 xx 10^(3) Am^(-1)`. The current required to be passed in a solenoid of length `10 cm` and number of turns `100`, so that the magnet gets demagnetized when inside the solenoid , is :A. `3A`B. `6A`C. `30mA`D. `60mA`

The coercivity of a certain permanent magnet is `4*0xx10^4Am^-1`. This magnet is placed inside a solenoid `15cm` long and having `600turns`. Calculate the current required to demagnetise the magnet completely.

The magnetic field at a point near the centre but outside a current carrying solenoid is zero. Explain why?

What is the magnitude of force on a wire of length `0*04m` placed inside a solenoid near its centre, making an angle of `30^@` with its axis? The wire carries a current of `12A` and the magnetic field due to the solenoid is of magnitude `0*25T`.

What is the potential energy of a dipole when it is perpendicular to a magnetic field?

What does the torque do?

Two identical magnetic dipoles of magnetic moments `1*0Am^2` each are placed at a separation of `2m` with their axes perpendicular to each other. What is the resultant magnetic field at a point midway between the dipoles?A. `5xx10^-7T`B. `sqrt5xx10^-7T`C. `10^-7T`D. `2xx10^-7T`

A long straight wire carries a current of `2A`. An electron travels with a velocity of `4xx10^4ms^-1` parallel to the wire `01mm` from it, and in a direction opposite to the current. What force does the magnetic field of current exert on the moving electron. Charge on electron `=16xx10^(-19)C`.

What is the magnitude of force on a wire of length `004m` placed inside a solenoid near its centre, making an angle of `30^@` with its axis? The wire carries a current of `12A` and the magnetic field due to the solenoid is of magnitude `025T`.