624 + Interview Questions in MAGNETIC-EFFECT-OF-CURRENT IN GENERAL AWARENESS Page 10 InterviewSolution

451.	The electron in a hydrogen atom circles around the proton with a speed of `218xx10^6ms^-1` in an orbit of radius `53xx10^-11m`. Calcualte (a) the equivalent current (b) magnetic field produced at the proton. Given charge on electron `1*6xx10^-19C` and `mu_0=4pixx10^-7TmA^-1`.
Answer» Here, `v=218xx10^6ms^-1`, `r=53xx10^-11m`, `e=16xx10^-19C`. (a) Time period of revolution of electron is given by, `T=(2pir)/(v)=(2pixx53xx10^-11)/(218xx10^6)=1528xx10^-16s` Equivalent current, `I=(charg e)/(time)=e/T` `=(16xx10^-19)/(1528xx10^-16)=105xx10^-3A` (b) `B=(mu_0)/(4pi)(2piI)/(r)=(10^-7xx2pixx105xx10^-3)/(53xx10^-11)` `=124T`

Discussion

452.	A horizontal circular loop carries a current that looks anticlockwise then viewed from above. It is replaced by an equivalent magnetic dipole NS, which of the following is true?A. The line NS should be along a diameter of the loop.B. The line NS should be perpendicular to the plane of the loop.C. South pole should be below the loop.D. North pole should be below the loop.
Answer» Correct Answer - B::C The magnetic field due to the horizontal circular loop is perpendicular to the plane of the loop and upwards. Therefore, the line NS of dipole should be perpendicular to the plane of the loop and north pole should be upwards or south pole should be downwards.

Discussion

453.	Two parallel wires in free spaces are `10cm` apart and each carries a current of `10A` in the same direction. The force one wire exerts on the other per metre of length isA. `2xx10^(-4)N` attractiveB. `2xx10^(-4)N`, repulsiveC. `2xx10^(-7)N` attractiveD. `2xx10^(-7)N`, repulsive
Answer» Correct Answer - A

Discussion

454.	A circular coil of `20turns` and radius `10cm` carries a current of `5A`. It is placed in a uniform magnetic field of `0*10T`. Find the torque acting on the coil when the magnetic field is applied (a) normal to the plane of the coil (b) in the plane of coil. Also find out the total force acting on the coil.
Answer» Here, `n=20`, `r=010m`, `I=5A`, `B=010T` Area of each turn of coil, `A=pir^2=22/7xx(01)^2=00314m^2` Torque acting on the coil, `tau=nIAB sin alpha` (a) Here, `alpha=0^@` so `tau=20xx5xx(00314)xx01xxsin0^@=0` (b) Here, `alpha=90^@` so `tau=20xx5xx(00314)xx01xxsin90^@=0*314N-m`. Coil carrying current acts as a magnetic dipole. The force acting on magnetic dipole in a uniform magnetic field is zero.

Discussion

455.	A short bar magnet of magnetic moment `09JT^-1` is placed with its axis at `30^@` to a uniform magnetic field. It experiences a torque of `0063J`. Calculate the magnitude of magnetic field. In which orientation will the bar magnet be in stable equilibrium in the magnetic field?
Answer» Here, `M=0JT^-1, theta=30^@` `tau=0063J, B=?` `tau=MB sin theta, B=(tau)/(Msintheta)=(0063)/(09xxsin30^@)` `=0*14T` For stable equilibrium, the magnetic moment `vecM` of magnet must be parallel to magnetic field `vecB`.

Discussion

456.	In a chamber, a uniform magnetic field of `65G(1G=10^-4T)` is maintained. An electron is shot into the field with a speed of `48xx10^6ms^-1` normal to the field. (i) Explain why the path of the electron is a circle. Determine the radius of the circular orbit. `(e=16xx10^(-19)C, m_e=91xx10^(-31)kg)`.
Answer» Here, `B=65xx10^-4T`, `v=48xx10^6m//s`, `e=16xx10^(-19)C`, `theta=90^@`, `m=91xx10^(-31)kg`, `r=?` (i) Force on the moving electron due to magnetic field will be, `F=evBsin theta`. The direction of this force is perpendicular to `vecv` as well as `vecB` therefore this force will only change the direction of motion of the electron without affecting its velocity i.e. this force will provide the centripetal force to the moving electron and hence, the electron will move on the circular path. If r is the radius of circular path traced by electron, then `evBsin90^@=mv^2//r` or `r=(mv)/(Be)=((91xx10^(-31))xx(48xx10^6))/((65xx10^-4)xx(16xx10^(-19)))=42xx10^-2m=42cm`

Discussion

457.	An electron of mass `090xx10^-30kg`, under the action of magnetic field moves in a circle of radius `20cm` at a speed of `30xx10^6ms^-1`. If a proton of mass `18xx10^-27kg` were to move in a cirlce of the same radius in the same magnetic field, find its speed.
Answer» Here, `m_e=9xx10^-31kg`, `r_e=002m`, `v_e=3xx10^6ms^-1`, `m_p=18xx10^-27kg`, `v_p=?`, `r_p=002m`. We know that `qvB=mv^2//r` or `v=qrB//m`, So `v_p/v_e=(q_pr_pB//m_p)/(q_er_eB//m_e)=(m_e)/(m_p)` [as `q_p=q_e`, `r_p=r_e`] `:. v_p=v_exxm_e/m_p=(3xx10^6xx9xx10^-31)/(18xx10^-27)=1*5xx10^3ms^-1`

Discussion

458.	The electron of hydrogen atom moves along a circular path of radius `05xx10^(-10)m` (i) with a speed of `40xx10^6ms^-1`. (ii) with a frequency `68xx10^(15)Hz`. Calculate the magnetic field produced at the centre of the circular path. `(e=16xx10^(-19)C)`.
Answer» Correct Answer - `256T`, `134T` (i) `T=2pi r//v` and `i=e/T=(ev)/(2pir)` (ii) `i=ev` Now use, `B=(mu_0)/(4pi)(2pi i)/(r)=(mu_0)/(4pi)(2pi)/(r)ev`

Discussion

459.	A loop of wire having 50 turns carries a current of `10A` in anticlockwise direction. If diameter of loop is `10cm`, what is the magnitude and direction of magnetic moment of current loop?
Answer» Here, `N=50, I=10A`, `r=10/2cm=5xx10^-2m` `M=? From M=NIA=NI(pir^2)` `M=50xx10xx314(5xx10^-2)^2` `M=3925Am^2` According to right hand thumb rule, direction of `vecM` is perpendicular to the plane of loop and away from observer.

Discussion

460.	An electron enters with a velocity `vecv=v_0hati` into a cubical region (faces parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in plane parallel to the x-y plane. Suggest a configuration of fields `vecE` and `vecB` that can lead to it.
Answer» As per question, the orbit of electron is spiral down inside the cube in a plane parallel to the x-y plane, therefore, `vecB` must be along `+z` direction, i.e., `vecB=B_0hatk`. Here the electric field should be along x-axis, i.e., `vecE=E_0hati`, where `E_0gt0`.

Discussion

461.	A length L of a wire carries a current I. Show that if the wire is formed into a circular coil, the maximum torque in a given magnetic fiel B, is developed when the coil has one turn only and the maximum torque has the value, `tau_(max)=L^2IB//4pi`.
Answer» Torque on the current loop is `tau=NIAB sin alpha` …(i) If there are N turns of circular coil, each of radius r, then `L=2pirN` or `r=L//2piN` Area of the coil, `A=pir^2=piL^2//(4pi^2N^2)=L^2//(4piN^2)` Putting this value in (i),we get `tau=NI[(L^2)/(4piN^2)]B sin alpha=(L^2IBsinalpha)/(4piN)`...(ii) From (ii), it is clear that `tau` is maximum if N is minimum, i.e., the number of turns `N=1`. Torque is maximum if `sin alpha=1` and `N=1` `:. tau_(max)=L^2IB//4pi`.

Discussion

462.	Equal current I flows in two segments of a circular loop in the direction shown in figure. Radius of loop is R. What is the magnitude of magnetic field induciton at the centre of the loop?
Answer» Magnetic field induction at O due to current through arc ADB is `B_1=(mu_0)/(4pi)I/Rtheta` Its direction is perpendicular to the plane of circular loop and upwards. Magnetic field induction at O due to current through arc ACB is `B_2=(mu_0)/(4pi)I/R(2pi-theta)` Its direction is perpendicular to the plane of circular loop and downwards. `:.` Net magnetic field induction at O is downwards, perpendicular to plane of loop `B=B_2-B_1=(mu_0)/(4pi)I/R[(2pi-theta)-theta]=(mu_0)/(4pi)(2I)/(R)(pi-theta)=(mu_0I(pi-theta))/(2piR)`

Discussion

463.	A short bar magnet placed with its axis at `30^@` with a uniform external magnetic field of `025T` experiences a torque of magnitude equal to `45xx10^-2J`. What is the magnitude of magnetic moment of the magnet?
Answer» Here, `theta=30^@`, `B=025T`, `tau=45xx10^-2J`, `M=?` As, `tau=MB sin theta` `:. M=(tau)/(B sin theta)=(45xx10^-2)/(025sin30^@)=0*36JT^-1`

Discussion

464.	A long straight wire carrying a current of `20A` is placed in an external uniform magnetic field of `3xx10^-4T` parallel to the current. Find the magnitude of resultant field at a point `2*0cm` away from the wire.
Answer» Here, `I=20A`, `B_2=3xx10^-4T`, `r=20xx10^-2m` Magnetic field due to a straight wire carrying current is `B_1=(mu_0)/(4pi)(2I)/(r)=10^-7xx(2xx20)/(20xx10^-2)=2xx10^-4T` This magnetic field will act perpendicular to magnetic field `B_2(=3xx10^-4T)`. Therefore, the magnitude of the resultant magnetic field `B=sqrt(B_1^2+B_2^2)=sqrt((2xx10^-4)^2+(3xx10^-4)^2)` `=3*6xx10^-4T`

Discussion

465.	In a chambe of a uniform magnetic field of `80G[1G=10^-4T]` is maintained. An electron with a speed of `40xx10^6ms^-1` enters the chamber in a direction normal to the field (i) Describe the path of electron. (ii) What is the frequency of revolution of electron? (iii) What happens to the path of the electron if it progressively loses its energy due to collisions with the atoms or molecules of the environment?
Answer» Here, `B=80G=80xx10^-4T`, `v=40xx10^6ms^-1`, `theta=90^@`. If r is the radius of circular path, then `Bevsin90^@=(mv^2)/(r)` or `r=(mv)/(Be)` `:. r=((91xx10^(-31))xx(40xx10^6))/((80xx10^-4)xx(16xx10^-19))` `=28xx10^-2m=28cm` The sense of rotation of electron in a circular path can be predicted from the direction of centripetal force, `vecF=-e(vecvxxvecB)`. If we see along the direction of `vecB`, the electron will be revolving in clockwise direction. (ii) The frequency of revolution of electron `v=(eB)/(2pim)=((16xx10^(-19))xx(80xx10^-4))/(2xx314xx(9*1xx10^(-31))` (iii) If the electron loses its energy in successive collisions, then electron loses its speed progressively. If after collision, the velocity of electron remains in the same plane of the initial circular orbit, the radius of the circualr orbit will decrease in proportion to the decreased speed. If it is not so then the path of the electron will be helical between two collisions.

Discussion

466.	Calculate the torque on a `200` turns rectangular coil of length `20cm` and breadth `10cm` carrying a current of `10A`, when placed in a magnetic field. The plane of the coil is making an angle of `60^@` with a magnetic field of `4T`.
Answer» Correct Answer - `80N-m` Here, `n=200`, `l=20cm=020m`, `b=10cm=010m` `B=4T`, `A=lxxb=020xx010=002m^2`, `I=10A`. Angle between `vecB` and normal on the plane of the coil `theta=90^@-60^@=30^@` Torque, `tau=nIBA sin theta` `=200xx10xx4xx002xxsin30^@=80N-m`

Discussion

467.	A charge q moving along the X-axis with a velocity `vecv` is subjected to a uniform magnetic field B acting along the Z-axis as it crosses the origin O. Figure. (i) Trace its trajectory . (ii) Does the charge gain kinetic energy as it enters the magnetic field? Justify yous answer.
Answer» (i) The trajectory of charge q will be a circle in Y-Z plane as shown in figure. (ii) The speed and kinetic energy of the charged particle remain constant, but velocity of the charged particle changes in direction.

Discussion

468.	What is the value of net force acting on a current carrying (i) rectangular coil and (ii) circular coil of same area placed in a uniform magnetic field? What will be the torque acting in each case?
Answer» The net force acting on each current carrying coil placed in magnetic field is zero. Torque on each coil is same provided the plane of each coil is equally inclined with the direction of magnetic field.

Discussion

469.	When a charged particle moving with a velocity `vecv` is subjected to a magnetic field `vecB`, the force acting on it is non zero. Would the particle gain any energy?
Answer» The magnetic force on a moving charged particle in a magnetic field is, `vecF=q(vecvxxvecB)`. Its direction is perpendicular to direction of motion of particle as well as perpendicular to the direction of magnetic field. Due to it, no work is done by the magnetic force on the charged particle. Therefore, the particle does not gain any kinetic energy.

Discussion

470.	A rectangular coil of sides `8cm` and `6cm` haivng 2000 turns and carrying a current of `200mA` is placed in a uniform magnetic field of `0*2T` directed along the positive x-axis. (a) What is the maximum torque the coil can experience? In which orientation does it experience the maximum Torque? (b) For which orientations of the coil is the torque zero? When is this equilibrium stable and unstable?
Answer» We know that a current loop, having n turns, each of area A, carrying current I, when placed in a magnetic field `vecB`, experience a torque whose magnitude is given by `tau=nIAB sin alpha` …(i) where `alpha` is the angle which the normal on the plane of teh current loop makes with the direction of magnetic field , i.e., angle between `vecA` and `vecB`. Here, `n=2000`, `I=200mA=200xx10^-3A`, `A=8xx6sq.cm=48xx10^-4m^2`. `B=02T`. (a) Torque acting on the coil will be maximum when `sin alpha=1` or `alpha=90^@` `:.` Maximum Torque, `tau_(max)=nIAB` `=2000xx(200xx10^-3)xx(48xx10^-4)xx02` `=0*384N-m` In this situation, the plane of the coil is parallel to the direction of magnetic field i.e., the plane of the coil is in the direction of X-axis. (b) Torque on the coil will be zero, if `sinalpha=0` or `alpha=0^@` or `180^@`. It will be so if plane of the coil is perpendicular to the direction of magnetic field. i.e., the plane of the coil is along Y or Z axis. The coil will be in stable equilibrium when `vecA` is parallel to `vecB` and in unstable equilibrium when `vecA` is antiparallel to `vecB`.

Discussion

471.	A rectangular coil of n turns each of area A, carrying current I, when suspended in a uniform magnetic field B, experiences a torque `tau=nIBAsin theta` where `theta` is the angle with a normal drawn on the plane of coil makes with the direction of magnetic field. This torque tends to rotate the coil and bring it in equilibrium position. In stable equilibrium state, the resultant force on the coil is zero. The torque on coil is also zero and the coil has minimum potential energy. Read the above passage and answer the following questions: (i) In which position, a current carrying coil suspended in a uniform magnetic field experiences (a) minimum torque and (b) maximum torque? (ii) A circular coil of 200 turns, radius `5cm` carries a current of `20A`. It is suspended vertically in a uniform horizontal magnetic field of `020T`, with the plane of the coil making an angle of `60^@` with the field lines. Calculate the magnitude of the torque that must be applied on it to prevent it from turning. (iii) What is the basic value displayed by the above study?
Answer» (i) As `tau=nIBA sin theta`, therefore, (i) `tau=0`, when `sin theta=0` or `theta=0^@`, i.e., when the plane of coil is perpendicular to the direction of magnetic field. (ii) `tau=`maximum, when `sin theta`=maximum `=1` or `theta=90^@` `tau_(max)=nIBAxx1=nIBA` It will be so when the plane of coil is parallel to the direction of magnetic field. (ii) Here, `n=200, r=005m, I=20A, B=020T, theta=90^@-60^@=30^@`. `tau=nIBA sin theta=nIB(pir^2)sin theta=200xx20xx020[(22//7)xx(005)^2]xxsin 30^@` `=0314N-m=031Nm`. (iii) From the above study, we find that when potential energy of the coil is minimum, both force and torque acting on the coil are zero. The same is ture in real life. They say "Nanak neevan jo chale, lage na tatti va", i.e., a person who is humble and boats of nothing, would be a happy person, with no pulls and pressures of life.

Discussion

472.	A long straight wire carrying a current of `25A` is placed in an external uniform magnetic field `30xx10^-4T` parallel to the current. Find the magnitude of the resultant magnetic field at a point `15cm` away from the wire.
Answer» Correct Answer - `448xx10^-4T` Here, `I=25A`, `r=15xx10^-2m`, `B_2=3xx10^-4T` Magnetic field due to straight current carrying wire, `B_1=(mu_0)/(4pi)(2I)/(r)=10^-7xx(2xx25)/((15xx10^-2))=10/3xx10^-4T` This magnetic field will act perpendicular to external magnetic field `B_2(=3xx10^-4T)`. Therefore, the magnitude of the resultant magnetic field is `B=sqrt(B_1^2+B_2^2)=sqrt((10/3xx10^-4)^2+(3xx10^-4)^2)` `=448xx10^-4T`

Discussion

473.	A circular coil of 16 turns and radius 10cm carrying a current of 0.75A rests with its plane normal to an external field of magnitude `50xx10^-2T`. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of `20s^-1`. What is the moment of inertia of the coil about its axis of rotation?
Answer» Here, `n=16`, `r=10cm=01m`, `i=0.75A`, `B=50xx10^-2T`, `v=2.0s^-1`, `I=?` `M=niA=nipir^2=16xx075xx22/7(01)^2=0377JT^-1` As `v=(1)/(2pi)sqrt((MB)/(I))` where I is the moment of inertia of the coil. `:. v^2=(MB)/(4pi^2I)` or `I=(MB)/(4pi^2v^2)=(0.377xx5.0xx10^-2)/(4xx(22/7)^2xx2^2)=12xx10^-4kgm^2`

Discussion

474.	A rectangular coil of sides l and b carrying a current I is subjected to a uniform magnetic field `vecB` acting at an angle `theta` to its plane. Write the expression for the torque acting on it. In which orientation of the coil in the magnetic field, the torque is (i) minimum and (ii) maximum.
Answer» Torque on the coil, `tau=IlB b cos theta=IlbB cos theta`. (i) Torque is minimum if `cos theta` is minimum, i.e., `cos theta=0` or `theta=90^@` i.e., the plane of coil is set perpendicular to the direction of magnetic field. (ii) Torque is maximum if `cos theta=1` or `theta=0^@`. It means the plane of coil is parallel to the direction of magnetic field.

Discussion

475.	Consider a wire carrying a steady current, I placed in a uniform magnetic field `vecB` perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do not work. This implies thatA. motion of charges inside the conductor is unaffected by `vecB` since they do not absorbB. Some charges inside the wire move to the surface as a result of `vecB`C. if the wire moves under the influence of `vecB`, no work is done by the forceD. if the wire moves under the influence of `vecB`, no work is done by the magnetic force on the ions, assumed fixed within the wire
Answer» Correct Answer - B::D (a) Motion of charges inside the conductor is affected by magnetic field `vecB`, due to magnetic force `vecF`, given by `vecF=q(vecvxxvecB)` (b) Due to magnetic force, some charges inside the wire move to the surface of wire. (c) The force on wire of length l, carrying current I when subjected to magnetic field `vecB` is, `vecF=I(veclxxvecB)`. It acts perpendicular to the plane containing `vecl` and `vecB` and is directed as given by Right Hand rule. If the wire moves under the influence of `vecB` at an angle `theta`, where `theta=90^@`, then work done, `W=Fscostheta`, can not be zero. (d) When wire moves under the influence of `vecB`, then displancement of the ions is perpendicular to the magnetic force `vecF`. Therefore work done is zero.

Discussion

476.	A long straight wire carrying current of `25A` rests on a table as shown in figure. Another wire PQ of length 1m, mass `2*5g` carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise?
Answer» Here, `I_1=25A, I_2=25A, l=1m, m=25g=25xx10^-3kg` In equilibrium position `mg=(mu_0)/(4pi)(2I_1I_2l)/(h)` or `h=(mu_0)/(4pi)(2I_1I_2l)/(mg)=(10^-7xx2xx25xx25xx1)/((2.5xx10^-3)xx98)=51xx10^-4m=051cm`

Discussion

477.	A multirange current meter can be constructed by using a galvanometer circuit as shown in figure. We want a current meter that can measure `10mA`, `100mA` and `1A` using a galvanometer of resistance `10Omega` and that produces maximum deflection for current of `1mA`. Find `S_1, S_2` and `S_3` that have to be used.
Answer» Here, `G=10Omega`, `I_g=1mA=10^-3A`. To convert a galvanometer into an ammeter of given range of current 0 to I, the shunt resistance required is `S=(I_gG)/(I-I_g)`. Case (i), `I=10mA=10xx10^-3A`, `S=S_1+S_2+S_3`. `:. S_1+S_2+S_3=(1mAxx10Omega)/((10-1)mA)=10/9Omega` Case (ii), When `I=100mA`, `S=S_2+S_3`, galvanometer resistance `=G+S_1` `:. S_2+S_3=(I_(g)(G+S_1))/(I-I_g)=(1mAxx(10+S_1))/((100-1)mA)=(10+S_1)/(99)` ...(ii) Case (iii), When `I=1A`, `S=S_3`, galvanometer resistance`=(G+S_1+S_2)` `:. S_3=(I_g(G+S_1+S_2))/(I-I_g)=(1mA[10+S_1+S_2])/((1000-1)mA)=(10+S_1+S_2)/(999)`...(iii) Putting the value of (ii) in (i), we get `S_1+(10+S_1)/(99)=10/9` or `S_1(1+1/99)=10/9-10/99=100/99` `:. S_1xx100/99=100/99` or `S_1=1Omega` From (ii), `S_2+S_3=1/99(10+1)=1/9` ...(iv) From (iii), `S_3=(10+1+S_2)/(999)=(11+S_2)/(999)` or `S_3-S_2/999=11/999` ...(v) Subtracting (v) from (iv), we get, `S_2=(S_2)/(999)=1/9-11/999=100/999` or `(S_2xx1000)/(999)=100/999` or `S_2=1/10=01Omega` From (iv), `1/10+S_3=1/9` or `S_3=1/9-1/10=1/90` or `S_3=0011Omega`

Discussion

478.	A magnet of length `2l` and pole strength m is divided in two equal parts along its length. What is magnetic moment of each part?
Answer» Pole strength(m) is halved. Length `(2l)` remains the same. As `M=mxx2l`, therefore, M becomes half.

Discussion

479.	Write the formula of magnetic moment of a current loop.
Answer» `M=NIA`, when N is number of turns I, the current and A=area of cross section of loop.

Discussion

480.	A small coil of `N` turns has an effective area `A` and carries a current `I`. It is suspended in a horizontal magnetic field `vecB` such that its plane is perpendicular to `vecB`. The work done in rotating it by `180^(@)` about the vertical axis isA. `NAlB`B. `2NAlB`C. `2piNAIB`D. `4piNAIB`
Answer» Correct Answer - B

Discussion

481.	In a permanent magnet at room temperature.A. magnetic moment of each molecule is zeroB. the individual molecules have non-zero magnetic moment which are all perfectly alignedC. domains are partially alignedD. domains are all perfectly algined
Answer» Correct Answer - C In a permanent magnet at room temperature domains of a magnet are partially aligned due to thermal agitation.

Discussion

482.	A proton enters a magnetic fiedl of flux density `1.5"weber"//m^(2)` with a velocity of `2xx10^(-7)m//sec` at an angle of `30^(@)` with the field. The force on the proton will beA. `2.4xx10^(-12)N`B. `0.24xx10^(-12)N`C. `24xx10^(-12)N`D. `0.02xx10^(-12)N`
Answer» Correct Answer - A

Discussion

483.	A beam of electrons passes underfected throgh unifromly perpendicular electric and magnetic fields. If the electric fiedl is swiched off, and the same magnetic field is mainted fiedl is maintetained the electrons move:A. in an elliptical orbitB. in acircular orbitC. along aparobolic pathD. along a straight line
Answer» Correct Answer - b

Discussion

484.	A uniform constant magnetic field `B` is directed at an angle of `45^(@)` to the `x axis` in the ` xy`- plane . ` PQRS` is a rigid, square wire frame carrying a steady current `I_(0)`, with its centre at the origin `O`. At time ` t = 0`, the frame is at rest in the position as shown in figure , with its sides parallel to the ` x and y` axis. Each side of the frame is of mass `M` and length `L`. (a) What is the torque `tau` about `O` acting on the frame due to the magnetic field? (b) Find the angle by which the frame rotates under the action of this torque in a short interval of time `Deltat`, and the axis about this rotation occurs .`( Deltat is so short that any variation in the torque during this interval may be neglected .) Given : the moment of interia of the frame about an axis through its centre perpendicular to its about an axis through its centre perpendicular to its plane is `(4)/(3) ML^(2)`. A. `SQ`B. `PQ`C. `PS`D. `PR`
Answer» Correct Answer - `A`

Discussion

485.	A uniform constant magnetic field `B` is directed at an angle of `45^(@)` to the `x axis` in the ` xy`- plane . ` PQRS` is a rigid, square wire frame carrying a steady current `I_(0)`, with its centre at the origin `O`. At time ` t = 0`, the frame is at rest in the position as shown in figure , with its sides parallel to the ` x and y` axis. Each side of the frame is of mass `M` and length `L`. (a) What is the torque `tau` about `O` acting on the frame due to the magnetic field? (b) Find the angle by which the frame rotates under the action of this torque in a short interval of time `Deltat`, and the axis about this rotation occurs .`( Deltat is so short that any variation in the torque during this interval may be neglected .) Given : the moment of interia of the frame about an axis through its centre perpendicular to its about an axis through its centre perpendicular to its plane is `(4)/(3) ML^(2)`. A. `(I_(0)L^(2)B)/(sqrt2) (hati + hatj)`B. `(I_(0)L^(2)B)/(sqrt2) (-hati + hatj)`C. `(I_(0)L^(2)B)/(sqrt2) (-hati - hatj)`D. `(I_(0)L^(2)B)/(sqrt2) (hati + hatj)`
Answer» Correct Answer - `B`

Discussion

486.	A uniform constant magnetic field `B` is directed at an angle of `45^(@)` to the `x axis` in the ` xy`- plane . ` PQRS` is a rigid, square wire frame carrying a steady current `I_(0)`, with its centre at the origin `O`. At time ` t = 0`, the frame is at rest in the position as shown in figure , with its sides parallel to the ` x and y` axis. Each side of the frame is of mass `M` and length `L`. (a) What is the torque `tau` about `O` acting on the frame due to the magnetic field? (b) Find the angle by which the frame rotates under the action of this torque in a short interval of time `Deltat`, and the axis about this rotation occurs .`( Deltat is so short that any variation in the torque during this interval may be neglected .) Given : the moment of interia of the frame about an axis through its centre perpendicular to its about an axis through its centre perpendicular to its plane is `(4)/(3) ML^(2)`. A. `(3I_(0)B)/(4M)(Deltat)^(2)`B. `(4I_(0)B)/(M)(Deltat)^(2)`C. `(2I_(0)B)/(3M)(Deltat)^(2)`D. `(4I_(0)B)/(3M)(Deltat)^(2)`
Answer» Correct Answer - `A`

Discussion

487.	`OABC` is a current carrying square loop an electron is projected from the center of loop along its diagonal `AC` as shown. Unit vector in the direction of initial acceleration will be .A. `hatk`B. `-((hati+hatj)/(sqrt2))`C. `-hatk`D. `(hati+hatj)/(sqrt2)`
Answer» Correct Answer - B `oversetrarrF = -e (oversetrarrV xx oversetrarrB)` Force will be along `BO` or unit vector `= -(hati + hatj)/(sqrt2a` .

Discussion

488.	An electron is projected along the axis of a circular conductor carrying some current. Electron will experience forceA. Along the axisB. Perpendicular to the axisC. At an angle of 4o with axisD. No force experienced
Answer» Correct Answer - D

Discussion

489.	A stream of electrons is projected horizontally to the right. A straight conductor carrying a current is supported parallel to the electron steam and above it. If the current in the conductor is from left to right, what will be the effect on the electron stream?A. The electron stream will be pulled upwardB. The electron stream will be pulled downwardC. The electron stream will be retartedD. The electron beam will be speeded up towards the right
Answer» Correct Answer - B

Discussion

490.	The resistance of a pivoted type galvanometer is `8Omega` and current for full scale deflection on it is `001A`. This galvanometer is to be converted into an ammeter of `5A` range. The only shunt available is `002Omega`. Find the value of R to be connected in series with the galvanometer coil, figure.
Answer» Correct Answer - `198Omega` `I_g(G+R)=(I-I_g)S` or `001(8+R)=(5-001)002=499xx002` or `R=(4.99xx0.02)/(0.01)-8=9.98-8` `=1.98Omega`

Discussion

491.	Two particles , each of mass ` m` and charge `q`, are attached to the two ends of a light rigid rod of length ` 2 R` . The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod isA. `q/(2m)`B. `q/m`C. `(2q)/m`D. `q/(pm)`
Answer» Correct Answer - A

Discussion

492.	In a current carrying long solenoid, the field produced does not depend uponA. Number of turns per unit lengthB. Current flowingC. Radius of the solenoidD. All of the above three
Answer» Correct Answer - C

Discussion

493.	The expression for magnetic induction inside a solenoid of length `L` carrying a current `I` and having `N` number of turns isA. `(mu_(0))/(4pi)N/(LI)`B. `mu_(0)NI`C. `(mu_(0))/(4pi)NLI`D. `mu_(0)N/L I`
Answer» Correct Answer - D

Discussion

494.	The magnetic induction at any point due to a long straight wire carrying a current isA. Proportional to the distance from the wireB. Inversely proportional to the distance from wireC. Inversely proportional to the square of the distance from the wireD. Does not depend on distance
Answer» Correct Answer - B

Discussion

495.	Which of the following graphs shows the variation of magnetic induction B with distance r from a long wire carrying currentA. B. C. D.
Answer» Correct Answer - C

Discussion

496.	In a galvanometer there is a deflection of 10 divisions per `50 mA`. The internal resistance of the galvanometer is `60Omega`. If a shunt of `2*5Omega` is connected to the galvanometer and there are 50 divisions in all on the scale of galvanometer what maximum current can this galvanometer read?
Answer» Correct Answer - (a) Increasing the turns by 20% (b) Lesser Since the galvanometer has 50 divisions, so current for full scale deflection is `I_g=1/10xx50mA=5mA=5xx10^-3A`, `G=60Omega`, `S=25Omega`. Let I be the maximum current which a galvanometer can read when shunted with resistance S, then `I_g=(IS)/(G+S)` or `I=(I_g(G+S))/(S)` `=((5xx10^-3)(60+25))/(2*5)` `=125xx10^-3A` `=125mA`

Discussion

497.	A proton accelerated by a potential difference `500 KV` moves though a transverse field of `0.51 T` as shown in figure. The angle `theta` through which the proton deviates from the intial direction of its motion is A. `15^(@)`B. `30^(@)`C. `45^(@)`D. `60^(@)`
Answer» Correct Answer - B

Discussion

498.	The charge on a particle `Y` is double the charge on particle `X`.These two particles `X` and `Y` after being accelerated through the same potential difference enter a region of uniform magnetic field and describe circular paths of radii `R_(1)` and `R_(2)` respectively. The ratio of the mass of `X` to that of `Y` isA. `((2R_(1))/(R_(2)))^(2)`B. `((R_(1))/(2R_(2)))^(2)`C. `(R_(1)^(2))/(2R_(2)^(2))`D. `(2R_(1))/(R_(2))`
Answer» Correct Answer - C

Discussion

499.	An electron is accelerated by a potential difference of 12000 volts. It then enters a uniform magnetic field of `10^(-3)T` applied perpendicular to the path of electron. Find the radius of path. Given mass of electron `=9xx10^(-31)kg` and charge on electron `=1.6xx10^(-19)C`A. `36.7m`B. `36.7m`C. `3.67m`D. `3.67m`
Answer» Correct Answer - B

Discussion

500.	A particle with `10^(-11)` coulomb of charge and `10^(-7)kg` mass is moving wilth a velocity of `10^(8)m//s` along the `y`-axis. A uniform static magnetic field `B=0.5` Tesla is acting along the `x`-direction. The force on the particle isA. `5xx10^(-11)N` along `hati`B. `5xx10^(3)N` along `hatk`C. `5xx10^(-11)N` along `-hatj`D. `5xx10^(-4)N` along `-hatk`
Answer» Correct Answer - D

Discussion

Explore topic-wise InterviewSolutions in .

The electron in a hydrogen atom circles around the proton with a speed of `2*18xx10^6ms^-1` in an orbit of radius `5*3xx10^-11m`. Calcualte (a) the equivalent current (b) magnetic field produced at the proton. Given charge on electron `1*6xx10^-19C` and `mu_0=4pixx10^-7TmA^-1`.

Two parallel wires in free spaces are `10cm` apart and each carries a current of `10A` in the same direction. The force one wire exerts on the other per metre of length isA. `2xx10^(-4)N` attractiveB. `2xx10^(-4)N`, repulsiveC. `2xx10^(-7)N` attractiveD. `2xx10^(-7)N`, repulsive

A short bar magnet of magnetic moment `0*9JT^-1` is placed with its axis at `30^@` to a uniform magnetic field. It experiences a torque of `0*063J`. Calculate the magnitude of magnetic field. In which orientation will the bar magnet be in stable equilibrium in the magnetic field?

An electron of mass `0*90xx10^-30kg`, under the action of magnetic field moves in a circle of radius `2*0cm` at a speed of `3*0xx10^6ms^-1`. If a proton of mass `1*8xx10^-27kg` were to move in a cirlce of the same radius in the same magnetic field, find its speed.

The electron of hydrogen atom moves along a circular path of radius `0*5xx10^(-10)m` (i) with a speed of `4*0xx10^6ms^-1`. (ii) with a frequency `6*8xx10^(15)Hz`. Calculate the magnetic field produced at the centre of the circular path. `(e=1*6xx10^(-19)C)`.

A loop of wire having 50 turns carries a current of `10A` in anticlockwise direction. If diameter of loop is `10cm`, what is the magnitude and direction of magnetic moment of current loop?

A length L of a wire carries a current I. Show that if the wire is formed into a circular coil, the maximum torque in a given magnetic fiel B, is developed when the coil has one turn only and the maximum torque has the value, `tau_(max)=L^2IB//4pi`.

Equal current I flows in two segments of a circular loop in the direction shown in figure. Radius of loop is R. What is the magnitude of magnetic field induciton at the centre of the loop?

A short bar magnet placed with its axis at `30^@` with a uniform external magnetic field of `0*25T` experiences a torque of magnitude equal to `4*5xx10^-2J`. What is the magnitude of magnetic moment of the magnet?

A long straight wire carrying a current of `20A` is placed in an external uniform magnetic field of `3xx10^-4T` parallel to the current. Find the magnitude of resultant field at a point `2*0cm` away from the wire.

Calculate the torque on a `200` turns rectangular coil of length `20cm` and breadth `10cm` carrying a current of `10A`, when placed in a magnetic field. The plane of the coil is making an angle of `60^@` with a magnetic field of `4T`.

A charge q moving along the X-axis with a velocity `vecv` is subjected to a uniform magnetic field B acting along the Z-axis as it crosses the origin O. Figure. (i) Trace its trajectory . (ii) Does the charge gain kinetic energy as it enters the magnetic field? Justify yous answer.

What is the value of net force acting on a current carrying (i) rectangular coil and (ii) circular coil of same area placed in a uniform magnetic field? What will be the torque acting in each case?

When a charged particle moving with a velocity `vecv` is subjected to a magnetic field `vecB`, the force acting on it is non zero. Would the particle gain any energy?

A long straight wire carrying a current of `25A` is placed in an external uniform magnetic field `3*0xx10^-4T` parallel to the current. Find the magnitude of the resultant magnetic field at a point `1*5cm` away from the wire.

A rectangular coil of sides l and b carrying a current I is subjected to a uniform magnetic field `vecB` acting at an angle `theta` to its plane. Write the expression for the torque acting on it. In which orientation of the coil in the magnetic field, the torque is (i) minimum and (ii) maximum.

A long straight wire carrying current of `25A` rests on a table as shown in figure. Another wire PQ of length 1m, mass `2*5g` carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise?

A magnet of length `2l` and pole strength m is divided in two equal parts along its length. What is magnetic moment of each part?

Write the formula of magnetic moment of a current loop.

A small coil of `N` turns has an effective area `A` and carries a current `I`. It is suspended in a horizontal magnetic field `vecB` such that its plane is perpendicular to `vecB`. The work done in rotating it by `180^(@)` about the vertical axis isA. `NAlB`B. `2NAlB`C. `2piNAIB`D. `4piNAIB`

In a permanent magnet at room temperature.A. magnetic moment of each molecule is zeroB. the individual molecules have non-zero magnetic moment which are all perfectly alignedC. domains are partially alignedD. domains are all perfectly algined

A proton enters a magnetic fiedl of flux density `1.5"weber"//m^(2)` with a velocity of `2xx10^(-7)m//sec` at an angle of `30^(@)` with the field. The force on the proton will beA. `2.4xx10^(-12)N`B. `0.24xx10^(-12)N`C. `24xx10^(-12)N`D. `0.02xx10^(-12)N`

`OABC` is a current carrying square loop an electron is projected from the center of loop along its diagonal `AC` as shown. Unit vector in the direction of initial acceleration will be .A. `hatk`B. `-((hati+hatj)/(sqrt2))`C. `-hatk`D. `(hati+hatj)/(sqrt2)`

An electron is projected along the axis of a circular conductor carrying some current. Electron will experience forceA. Along the axisB. Perpendicular to the axisC. At an angle of 4o with axisD. No force experienced

In a current carrying long solenoid, the field produced does not depend uponA. Number of turns per unit lengthB. Current flowingC. Radius of the solenoidD. All of the above three

The expression for magnetic induction inside a solenoid of length `L` carrying a current `I` and having `N` number of turns isA. `(mu_(0))/(4pi)N/(LI)`B. `mu_(0)NI`C. `(mu_(0))/(4pi)NLI`D. `mu_(0)N/L I`

The magnetic induction at any point due to a long straight wire carrying a current isA. Proportional to the distance from the wireB. Inversely proportional to the distance from wireC. Inversely proportional to the square of the distance from the wireD. Does not depend on distance

Which of the following graphs shows the variation of magnetic induction B with distance r from a long wire carrying currentA. B. C. D.

In a galvanometer there is a deflection of 10 divisions per `50 mA`. The internal resistance of the galvanometer is `60Omega`. If a shunt of `2*5Omega` is connected to the galvanometer and there are 50 divisions in all on the scale of galvanometer what maximum current can this galvanometer read?

A proton accelerated by a potential difference `500 KV` moves though a transverse field of `0.51 T` as shown in figure. The angle `theta` through which the proton deviates from the intial direction of its motion is A. `15^(@)`B. `30^(@)`C. `45^(@)`D. `60^(@)`

The electron in a hydrogen atom circles around the proton with a speed of `218xx10^6ms^-1` in an orbit of radius `53xx10^-11m`. Calcualte (a) the equivalent current (b) magnetic field produced at the proton. Given charge on electron `1*6xx10^-19C` and `mu_0=4pixx10^-7TmA^-1`.

A short bar magnet of magnetic moment `09JT^-1` is placed with its axis at `30^@` to a uniform magnetic field. It experiences a torque of `0063J`. Calculate the magnitude of magnetic field. In which orientation will the bar magnet be in stable equilibrium in the magnetic field?

An electron of mass `090xx10^-30kg`, under the action of magnetic field moves in a circle of radius `20cm` at a speed of `30xx10^6ms^-1`. If a proton of mass `18xx10^-27kg` were to move in a cirlce of the same radius in the same magnetic field, find its speed.

The electron of hydrogen atom moves along a circular path of radius `05xx10^(-10)m` (i) with a speed of `40xx10^6ms^-1`. (ii) with a frequency `68xx10^(15)Hz`. Calculate the magnetic field produced at the centre of the circular path. `(e=16xx10^(-19)C)`.

A short bar magnet placed with its axis at `30^@` with a uniform external magnetic field of `025T` experiences a torque of magnitude equal to `45xx10^-2J`. What is the magnitude of magnetic moment of the magnet?

A long straight wire carrying a current of `25A` is placed in an external uniform magnetic field `30xx10^-4T` parallel to the current. Find the magnitude of the resultant magnetic field at a point `15cm` away from the wire.