InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 401. |
Due to the flow of current in a circular loop of radius `R`, the magnetic induction produced at the centre of the loop is `B`. The magnetic moment of the loop is (`mu_(0)`=permeability constant)A. `BR^(3)//2pi mu_(0)`B. `2piBR^(3)//mu_(0)`C. `BR^(2)//2pimu_(0)`D. `2piBR^(2)//mu_(0)` |
|
Answer» Correct Answer - B |
|
| 402. |
The magnetic moment of a circular coil carrying current isA. Directly proportional to the length of the wire in the coilB. Inversely proportional to the length of the wire in the coilC. Directly proportional to the square of the length of the wire inthe coilD. Inversely proportional to the square of the length of the wire inthe coil |
|
Answer» Correct Answer - C |
|
| 403. |
For the magnetic field to be maximum due to a small element of current carrying conductor at a point, the angle between the element and the line joining the element to the given point must beA. `0^(@)`B. `90^(@)`C. `180^(@)`D. `45^(@)` |
|
Answer» Correct Answer - B |
|
| 404. |
Same current i is flowing in the three infinitely long wires along positive x-,y- and z-directions. The magnetic filed at a point (0,0,-a) would be |
|
Answer» The point `(0, 0, -a)` lies on z-axis. Therefore the magnetic field induction at the given point due to current along z-axis is zero. The magnetic field induction due to current along x-axis at the given point is `vecB_x=(mu_0)/(4pi)I/ahatj` The magnetic field induction due to current along y-axis at the given point is `vecB_y=(mu_0)/(4pi)I/a(-hatj)` `:.` Total magnetic field induction, `vecB=vecB_x+vecB_y=(mu_0)/(4pi)I/a(hatj-hati)` |
|
| 405. |
Equal currents `i=1` A are flowing through the wires parallel to y-axis located at `x=+1m, x=+2m, x=+4m` and so on...., etc. but in opposite directions as shown in Fig The magnetic field (in tesla) at origin would be |
|
Answer» Correct Answer - `1*33xx10^-7hatk` Mag. field induction at O due to current through one wire is `vecB_1=(mu_0)/(4pi)(2I)/(r)hatk`, we have `vecB=(mu_0)/(4pi)2xx1[1/1-1/2+1/4-1/8+1/16+…]hatk` `=(mu_0)/(2pi)[(1+1/4+1/16+…)hatk-(1/2+1/8+…)hatk]` `=(mu_0)/(2pi)[((1)/(1-1/4))hatk-1/2((1)/(1-1/4))hatk]` `=(mu_0)/(2pi)2/3hatk=(mu_0)/(4pi)(4hatk)/(3)` `=10^-7xx4/3hatk=1*33xx10^-7hatk` |
|
| 406. |
Find the magnetic induction at the point O of the following figures if the wire carrying a current I = 15A has the shapes shown here. The radius of the curved part R=5cm, the linear parts of the wire are very long. |
|
Answer» Correct Answer - a) `B=(mu_(0)I)/(4piR)sqrt(pi^(2)+4)= 0.11 mT, b) B=`(mu_(0)I)/(4piR)sqrt(2+2pi+pi^(2))`= 0.13mT, (C ) B = `(mu_(0)I)/(8piR)sqrt(8 + 9pi^(2) = 0.15 mT |
|
| 407. |
A current I flows along a thin-walled, long, half-cylinder of radius R(figure) Find the magnetic induction at a point on the axis of the cylinder |
|
Answer» Correct Answer - `(mu_(0)I)/(pi^(2)R)` |
|
| 408. |
If induction of magnetic field at a point is `B` and energy density is `U` then which of the following graphs is correctA. B. C. D. |
|
Answer» Correct Answer - A |
|
| 409. |
An electron moves in a circular orbit with a uniform speed `v`.It produces a magnetic field `B` at the centre of the circle. The radius of the circle is proportional toA. `(B)/(v)`B. `(v)/(B)`C. `sqrt((v)/(B))`D. `sqrt((B)/(v))` |
|
Answer» Correct Answer - c |
|
| 410. |
An electron moves in a circular orbit with a uniform speed `v`.It produces a magnetic field `B` at the centre of the circle. The radius of the circle is proportional toA. `B/v`B. `v/R`C. `sqrt(v/B)`D. `sqrt(B/v)` |
|
Answer» Correct Answer - C |
|
| 411. |
An electric field of 1500 V/m and a magnetic field of 0.40 Wb/`m^(2)` act on a moving electron. The minimum uniform speed along a straight line, the electron could have isA. `1.6xx10^(15)m//s`B. `6xx10^(-16)m//s`C. `3.75xx10^(3)m//s`D. `3.75xx10^(2)m//s` |
|
Answer» Correct Answer - C |
|
| 412. |
At what distance from a long straight wire carrying current of 12A will be the magnetic field be the equal to `3xx10^(-5) (Wb)//(m^(2))` ?A. `8xx10^(-2)m`B. `12xx10^(-2)m`C. `18xx10^(-2)m`D. `24xx10^(-2)m` |
|
Answer» Correct Answer - a |
|
| 413. |
Write mathematical form of tangent law in magnetism. |
| Answer» `F=H tan theta`, where F and H are strengths of two magnetic fields acting perpendicular to each other and `theta` is angel which freely suspended magnet makes with the direction of H. | |
| 414. |
A bar magnet having a magnetic moment of `1.0 xx 10^4 J T^(-1)` is free to rotate in a horizontal plane. A horizontal magnetic field `B = 4 xx 10^(-5) T` exists in the space. Find the work done in rotating the magnet slowly from a direction parallel to the field to a direction `60^@` from the field. |
|
Answer» Here, `M=1*0xx10^4J//T`, `B=4xx10^-5T`, `W=?theta_1=0^@`, `theta_2=60^@` `W=-MB(costheta_2-costheta_1)=-1*0xx10^4xx4xx10^-5(cos60^@-cos60^@)=-0*4(1/2-1)=0*2J` |
|
| 415. |
A uniform magnetic field gets modified as shown in figure. When two specimens X and Y are placed in it. Identify the specimen X and Y. |
| Answer» X is diamagnetic as it expels magnetic lines Y is paramagnetic or ferromagnetic as it pulls the magnetic lines into it. | |
| 416. |
A charged particle of mass m and charge e is released from rest in an electric field E. If kinetic energy of particle after time t is `E_k` and linear momentum is p, thenA. `p=Eet`B. `p=2eEt`C. `E_K=(e^2E^2t^2)/(2m)`D. `E_K=(e^2E^2t^2)/(m)` |
|
Answer» Correct Answer - A::C (a) As `v=u+at`, `u=0` `v=at` As `a=(eE)/(m)` `:. v=(eEt)/(m)` and `p=mv` so `p=(meEt)/(m)=eEt` (c) `E_k=(P^2)/(2m)=(e^2E^2t^2)/(2m)` |
|
| 417. |
Establish analytically that the gain in kinetic energy of the charged particle moving in a magnetic field is zero. |
|
Answer» Total gain in K.E.=work done `=int vecF_m*dvecr = int_(t_(1))^(t_(2))vecF_m*(dvecr)/(dt)dt=int_(t_(1))^(t_(2))vecF_m*vecvdt` `int_(t_(1))^(t_(2)) q(vecvxxvecB)*vecvdt=int_(t_(1))^(t_(2))q(vecvxxvecv)*vecBdt=0` |
|
| 418. |
What is the magnetic effect of current ? Describe the nature of the magnetic field related with the current in circular coil. |
| Answer» When a current is passed through a conductor, magnetic field is produced around the conductor. It is called magnetic effect of current. The magnetic field is in the form of concentric circular magnetic lines of force for a linear conductor carrying current. The magnetic field is in the form of parallel straight lines at the centre and cocentric magnetic lines near the circular coil carrying current. | |
| 419. |
Figure shows a right-angled isosceles `DeltaPQR` having its base equal to a. A current of I ampere is passing downwards along a thin straight wire cutting the plane of paper normally as shown at Q. Likewise a similar wire carries an equal current passing normally upwards at R. Find the magnitude and direction of the magnetic field induction B at P. Assume the wires to be infinity long. |
|
Answer» Given `PQ=r_1`, `PR=r_2` In `DeltaPQR`, `a^2=r_1^2+r_2^2` Magnetic field induction at point P due to the current passing through wire at Q is `B_1=(mu_0I)/(2pir_1)` acting along PR Magnetic field induction at point P due to current passing through wire at R is `B_2=(mu_0I)/(2pir_2)` acting along PQ. Since `vecB_1` and `vecB_2` are perpendicular to each other, the resultant magnetic field induction at P is `B=sqrt(B_1^2+B_2^2)=(mu_0I)/(2pi)[(1/r_1)^2+(1/r_2)^2]^(1//2)` `=(mu_0I)/(2pi)[(r_1^2+r_2^2)/(r_1^2r_2^2)]^(1/2)=(mu_0I)/(2pi)((a^2)/(r_1^2r_2^2))^(1/2)=(mu_0Ia)/(2pir_1r_2)` |
|
| 420. |
Figure shows a right-angled isosceles `DeltaPQR` having its base equal to a. A current of I ampere is passing downwards along a thin straight wire cutting the plane of paper normally as shown at Q. Likewise a similar wire carries an equal current passing normally upwards at R. Find the magnitude and direction of the magnetic field induction B at P. Assume the wires to be infinity long. |
|
Answer» Here, `PQ=QR=r`. In art angled `DeltaPQR`, we have, `a^2=r^2+r^2=2r^2` or `r=a//sqrt2` Magnetic field at point P due to current through wire at Q is `B_1=(mu_0)/(4pi)(2I)/(r)=(mu_0I)/(2pir)=(mu_0I)/(2pi(a//sqrt2))=(mu_0I)/(sqrt2pia)` It is acting along PR. Magnetic field at point P due to current through wire at R is `B_2=(mu_0I)/(sqrt2pia)`, acting along PQ As `vecB_1` and `vecB_2` are acting perpendicular to each other, so the resultant magnetic field at P is `B=sqrt(B_1^2+B_2^2)=sqrt(((mu_0I)/(sqrtpia))^2+((mu_0I)/(sqrt2pia))^2)` `=sqrt(2)(mu_0I)/(sqrt2pia)=(mu_0I)/(pia)` It acts towards the mid-point of QR. |
|
| 421. |
An electron is moving vertically downwards. If it passes through a magnetic field which is directed from south to north in a horizontal plane, then in which direction the electron would be deflected? |
| Answer» Correct Answer - Towards west | |
| 422. |
An electron is moving vertically downwards. If it passes through a magnetic field which is directed from south to north in a horizontal plane, then in which direction the electron would be deflected?A. EastB. WestC. NorthD. South |
|
Answer» Correct Answer - A |
|
| 423. |
Magnetic field applied on a cyclotron is `0*7T` and radius of its dees is `1*8m`. What will be the energy of the emergent protons in `MeV`? Mass of proton`=1*67xx10^-27kg`. |
|
Answer» `E=(q^2B^2r^2)/(2m)=((1*6xx10^-19)^2xx(0*7)^2xx(1*8)^2)/(2xx1*67xx10^-27)J=((1*6xx10^-19)^2xx(0*7)^2xx(1*8)^2)/(2xx1*67xx10^-27xx1*6xx10^-13)MeV` `=76MeV` |
|
| 424. |
An alternate electric field of frequency v, is applied across the dees `(radius=R)` of a cyclotron that is being used to accelerate protons `("mass"=m)`. The operating magnetic field (b) used in the cyclotron and the kinetic energy (K) of the proton beam, produced by it, are given byA. `B=(mv)/(e)"and"K=2mpi^(2)v^(2)R^(2)`B. `B=(2pimv)/(e)"and"K=m^(2)pivR^(2)`C. `B=(2pimv)/(e)"and"K=2mpi^(2)v^(2)R^(2)`D. `B=(mv)/(e)"and"K=m^(2)pivR^(2)` |
|
Answer» Correct Answer - c |
|
| 425. |
A cyclotron has an oscillatory frequency of `10MHz` and a dee radius of `60cm`. Calculate the magnetic field required to accelerate the deutrons of mass `3*3xx10^(-27)kg` and charge `1*6xx10^(-19)C`. Find the energy of deutrons emerging from the cyclotron. |
|
Answer» Correct Answer - `1*3T`; `14*74MeV` Here, `v=10xx10^6Hz=10^7Hz`, `r=0*60m` `m=3*3xx10^(-27)kg`, `q=1*6xx10^(-19)C` As `v=(qB)/(2pim)`, so, `B=(2pimv)/(q)` `:. B=(2xx3*142xx(3*3xx10^(-27))xx10^7)/((1*6xx10^(-19)))=1*3T` Max. KE of the emerging deutron is `K_(max)=(q^2B^2r^2)/(2m)` `=((1*6xx10^(-19))xx(1*3)^2xx(0*6)^2)/(2xx(3*3xx10^(-27)))J` `=((1*6xx10^(-19))^2xx(1*3)^2xx(0*6)^2)/(2xx(3*3xx10^(-27))xx(1*6xx10^(-13)))=MeV` `=14*74MeV` |
|
| 426. |
An ammeter gives full scale deflection with a deflection for a current of `0*05A`. Calculate the length of shunt wire required to convert the galvanometer into an ammeter of range 0 to `5A`. The diameter of the shunt wire is `2mm` and its resistivity is `5xx10^-7Omegam`. |
|
Answer» Correct Answer - `3*174m` in parallel `S=(I_gG)/(I-I_g)=(1xxG)/(10-1)=G/9Omega`, Resistance of ammeter formed `R_p=(SxxG)/(G+S)=((G//9)xxG)/(G+G//9)=(G//9)/(10//9)=G/10` `:. R_p/S=(G//10)/(G//9)=9/10` |
|
| 427. |
Velocity and acceleration vector of a charged particle moving in a magnetic field at some instant are `vecv=3hati+4hatj and veca=2hati+xhatj`. Select the correct options.A. `x=-1*5`B. `x=3`C. magnetic field is along z-directionD. K.E. of particle is constant |
|
Answer» Correct Answer - A::C::D (a) Since magnetic force `(F_m)` is always perpendicular to the velocity vector `vecv`, i.e., `vecF_m_|_vecv` or `veca_|_vecv` or `veca.vecv=0` or `(2hati+xhatj).(3hati+4hatj)=0`, `x=-1*5`. (c) Since `vecB` is `_|_` to the plane containing the velocity vector, in this case (XY-plane, `vecB` is along Z-direction) (d) Since work done by magnetic force is zero, So K.E. of particle remains constant. |
|
| 428. |
A particle of mass `m` and charge `q` enters a magnetic field `B` perpendicularly with a velocity `v`, The radius of the circular path described by it will beA. `Bq//mv`B. `mq//Bv`C. `mB//qv`D. `mv//Bq` |
|
Answer» Correct Answer - D |
|
| 429. |
An electron moving towards the east enters a magnetic field directed towards the north. The force on the electron will be directedA. Vertically upwardB. Vertically downwardC. Towards the westD. Towards the south |
|
Answer» Correct Answer - B |
|
| 430. |
A circular coil of 100 turns and having an effective radius of `5cm` carries a current of `0*1` ampere. How much work is required to turn it in an external magnetic field of `1*5 weber//m^2` through `180^@` about an axis perpendicular to magnetic field. The plane of the coil is initially perpendicular to the magnetic field. |
|
Answer» Here, `N=100, r=5cm=0*05m`, `I=0*1amp., W=?, B=1*5Wb//m^2`, `theta_1=0^@, theta_2=180^@` Now, `area A =pir^2=3*14(0*05)^2sqm` `W=-MB(costheta_2-costheta_1)` `=-(NIA)B(costheta_2-costheta_1)` `=-100xx0*1xx3*14(0*05)^2xx1*5[cos180^@-cos0^@]` `=-31*4(0*05)^2xx1*5(-1,-1)=0*24J` |
|
| 431. |
Two identical circular coils, P and Q, carrying currents `1A` and `sqrt3A` respectively, are placed concentrically and perpendicular to each other lying in the XY and YZ plane. Find the magnitude and direction of the net magnetic field at the centre of the coils. |
|
Answer» Correct Answer - `(mu_0)/(R)T` in X-Z plane `B_P=(mu_0I_1)/(2R)=(mu_0xx1)/(2R)=(mu_0)/(2R)` acting along Z-axis `B_Q=(mu_0I_2)/(2R)=(mu_0xxsqrt3)/(2R)=sqrt3/2(mu_0)/(R)` acting along X-axis Net magnetic field is `B=sqrt(B_P^2+B_Q^2)=(mu_0)/(2R)[(1)^2+(sqrt3)^2]^(1//2)` `=(mu_0)/(R)T` (in X-Z plane) |
|
| 432. |
A circular coil of `100` turns and having a radius of `0*05m` carries a current of `0*1A`. Calculate the work required to turn the coil in an external magnetic field of `1*5T` through `180^@` about an axis perpendicular to the magnetic field. The plane of the coil is initially at right angles to the magnetic field. |
|
Answer» Correct Answer - `0*24J` Here, `n=100`, `r=0*05m`, `I=0*1A` `W=?`, `B=1*5T`, `theta_1=0^@`, (When plane of coil is at `90^@` to B) `theta_2=180^@` `M=nIA=nI(pir^2)` `=100xx0*1(3*14)(0*05)^2` `=7*85xx10^-2Am^2` `W=-MB(costheta_2-costheta_1)` `=-7*85xx10^-2xx1*5(cos180^@-cos0^@)` `W=0*24J` |
|
| 433. |
A long solenoid has 1000 turns per metre and carries a current of `1A`. It has a soft iron core of `mu_r=1000`. The core is heated beyond the Curie temperature, `T_c`.A. The `vecH` field in the solenoid is (nearly) unchanged but the `vecB` field decreases drasticallyB. The `vecH` and `vecB` fields in the solenoid are nearly unchangedC. The magnetisation in the core reverse directionD. The magnetisation in the core diminishes by a factor of about `10^8` |
|
Answer» Correct Answer - A::D In the solenoid, `H=nI=a` constant and `B=mu_0mu_rnI` changes due to variation in `mu_r`. When temperature of the iron core of solenoid (which is ferromagnetic materials) is raised beyond curie temperature, then soft iron core behaves as para magnetic material. We know that `(chi_m)_(para)~~10^-5.(chi_m)_(Ferro)~~10^3` `:. ((chi_m)_(Ferro))/((chi_m)_(Para))=10^3/10^-5=10^8` Thus magnetisation of the core diminishes by a factor `10^8`. Hence, options (a) and (d) are correct. |
|
| 434. |
The magnetic flux density applied in a cyclotron is `3*5T`. What will be the frequency of electric field that must be applied between the dees in order (a) to accelerate protons (b) `alpha`-particles? mass of proton `1*67xx10^(-27)kg`. |
|
Answer» (a) `v=(Bq)/(2pim)=(3*5xx1*6xx10^(-19))/(2xx(22//7)xx1*67xx10^(-27))=5*35xx10^7Hz`. (b) `V=(Bxx2e)/(2pixx4m)=(3*5xx2xx1*6xx10^-19)/(2xx(22/7)xx4xx1*67xx10^(-27))=2*675xx10^7Hz`. |
|
| 435. |
A circular loop of radius `0*2m` carrying a current of `1A` is placed in a uniform magnetic field of `0*5T`. The magnetic field is perpendicular to the plane of the loop. What is the force experienced by the loop? |
| Answer» The current carrying loop is equivalent to a magnetic dipole. The magnetic dipole does not experience any net force in a uniform magnetic field. | |
| 436. |
Two infinitely long parallel wires carry current `I_(1) = 8A` and `I_(2) = 10` A in opposite directions. The separation between the wires is `d = 0.12` m. Find the magnitude of magnetic field at a point P that is at a perpendicular distance `r_(1) = 0.16` m and `r_(2) = 0.20` m respectively from the wires. |
|
Answer» Correct Answer - `sqrt(40)muT` |
|
| 437. |
`1A` current flows through an infinitely long straight wire. The magnetic field produced at a point 1 metres away from it isA. `2xx10^(-3)` TeslaB. `2/10` TeslaC. `2xx10^(-7)` TeslaD. `2pixx10^(-6)` Tesla |
|
Answer» Correct Answer - C |
|
| 438. |
What current must flow in an infinitely long straight wire to produce a magnetic field of `4xx10^5T` at `8cm` away from the wire? |
|
Answer» Correct Answer - `16A` Here, `I=?, B=4xx10^-5T, r=8xx10^-2m` `B=(mu_0)/(4pi)(2I)/(r)` or `4xx10^-5=(10^-7xx2xxI)/(8xx10^-2)` or `I=(8xx4xx10^-7)/(2xx10^-7)=16A` |
|
| 439. |
An infinitely long, straight conductor `AB` is fixed and a current is passed through it. Another movable straight wire `CD` of finite length and carrying current is held perpendicular to it and released. Neglect weight of the wire A. The rod `CD` will move upwards parallel to itselfB. The rod `CD` will move downward parallel to itselfC. The rod `CD` will move upward and turn clockwise at the same timeD. The rod `CD` will move upward and turn anti –clockwise at the same time |
|
Answer» Correct Answer - C |
|
| 440. |
An infinitely long hollow conducting cylinder with inner radius `(r )/(2)` and outer radius `R` carries a uniform current ra density along its length . The magnitude of the magnetic field , ` | vec(B)|` as a function of the radial distance `r` from the axis is best represented byA. B. C. D. |
|
Answer» Correct Answer - `D` `B xx 2 pix = mu_(0)J(pi(x^(2)-(R^(2))/(4)))` `B =(mu_(0)J)/(2pi) (x-(R^(2))/(4x))` |
|
| 441. |
If the strength of the magnetic field produced 10cm away from a infinitely long straight conductor is`10^(-5)"Weber"//m^(2)` , the value of the current flowing in the conductor will beA. 5 ampereB. 10 ampereC. 500 ampereD. 1000 ampere |
|
Answer» Correct Answer - A |
|
| 442. |
Two short magnets of equal dipole moments `M` are fastened perpendicularly at their centres (figure). The magnitude of the magnetic field at a distance `d` from the centre on the bisector of the right angle is A. `(mu_0)/(4pi)(2Msqrt2)/(d^3)`B. `(mu_0)/(4pi)(M)/(d^3)`C. `(mu_0)/(4pi)(Msqrt2)/(d^3)`D. `(mu_0)/(4pi)(2M)/(d^3)` |
|
Answer» Net magnetic moment of the couple of magnets is `sqrt(M^2+M^2)=Msqrt2` at `45^@` (midway). Point P lies an axial line of net magnetic moment. Therefore, `B=(mu_0)/(4pi)(2Msqrt2)/(d^3)` |
|
| 443. |
If a long hollow copper pipe carriers a direct current, the magnetic field associated with the current will be:A. Only inside the pipeB. Only outside the pipeC. Neither inside nor outside the pipeD. Both inside and outside the pipe |
|
Answer» Correct Answer - B |
|
| 444. |
If a long hollow copper pipe carriers a direct current, the magnetic field associated with the current will be:A. Only inside the rodB. Only outside the rodC. Both inside and outside the rodD. Neither inside nor outside the rod |
|
Answer» Correct Answer - C |
|
| 445. |
If a long hollow copper pipe carriers a direct current, the magnetic field associated with the current will be:A. inside the pipe onlyB. outside the pipe onlyC. both inside and outside the pipeD. no where |
|
Answer» Correct Answer - b |
|
| 446. |
A circular loop carrying a current is replaced by an equivalent magnetic dipole. A point on the axis of the loop is inA. An end-on positionB. A broad side-on positionC. Both (a) and (b)D. Neither (a) nor (b) |
|
Answer» Correct Answer - A |
|
| 447. |
A 100 turn closely wound circular coil of radius `10cm` carries a current of `3*2A`. (i) What is the field at the centre of the coil? (ii) What is the magnetic moment of this arrangement? The coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of `2T` in the horizontal direction exists such that initially the axis of the coil is in the direction of the field. The coil rotates through an angle of `90^@` under the influence of the magnetic field. (iii) What are the magnitudes of the torques on the coil in the initial and final positions? (iv) What is the angular speed acquired by the coil when it has rotated by `90^@`? The moment of inertia of the coil is `0*1kgm^2`. |
|
Answer» (i)Here, `n=100`, `r=0*10m`, `i=3*2A`, `B=2T`, `I=0*1kgm^2` `B=(mu_0)/(4pi)(2pini)/(r)` `=10^-7xx2xx22/7xx(100xx3*2)/(0*10)=2xx10^-3T` (ii) `M=niA=nipir^2` `=100xx(3*2)xx(22/7)xx(0*10)^2=10Am^2` (iii) `tau=|vecMxxvecB|=MB sin theta` where `theta` is the angle between `vecM` and `vecB` or between `vecA` and `vecB`. Initially, `theta=0^@`, `tau=MB sin 0^@=0`. Finally, `theta=90^@`, `tau=MBsin90^@=MB` `=10xx2=20Nm` (iv) `tau=Ialpha=I(domega)/(dt)=I(domega)/(dt)xx(dtheta)/(dt)` `=I(domega)/(d theta)xxomega=MB sin theta`. `:. Iomegadomega=MB sin thetad theta`. Integrating it within the given conditions, `I int_(0)^(omega) omega d omega = int_(0)^(pi//2) MB sin theta d theta` `I(omega^2)/(2)=MB(-costheta)_0^(pi//2)` `=-MB[cospi/2-cos0^@]=MB` or `omega=((2MB)/(I))^(1//2)=((2xx20)/(0*1))^(1//2)` `=20rad//s` Change in KE of rotation=work done in rotation `1/2Iomega^2=MB(cos theta_1-cos theta_2)` where `theta_1=0^@`, `theta_2=90^@`, `I=0*1kgm^2`, `MB=20Nm`. `:. omega=[(2MB(cos theta_1-cos theta_2))/(I)]^(1//2)` `=[(2xx20xx(cos 0^@-cos90^@))/(0*1)]^(1//2)` `=20rad//s` |
|
| 448. |
A short bar magnet placed with its axis at `30^@` to a uniform magnetic field of `0*2T` experiences a torque of `0*06N-m`. Calculate magnetic moment of the magnet. What orientation of magnet corresponds to its stable equilibrium in the magnetic field? |
|
Answer» Here, `theta=30^@, B=0*2T`, `tau=0*06N-m, M=?` From `tau=Mbsintheta`, `M=(tau)/(B sin theta)=(0*06)/(0*2 sin 30^@)=0*6Am^2` Potential energy of magnetic dipole `U=-MB cos theta` In stable equilibrium, P.E. is minimum `:. cos theta=1` or `theta=0^@` i.e., bar magnet will be in stable equilibrium. when its magnetic moment `vecM` is parallel to magnetic field `vecB`. |
|
| 449. |
A proton (charge `=1*6xx10^(-19)C`, mass `m=1*67xx10^(-27)kg`) is shot with a speed `8xx10^6ms^-1` at an angle of `30^@` with the x-axis. A uniform magnetic field `B=0*30T` exists along the x-axis. Show that the path of the proton is helix and find the radius of the helix. |
|
Answer» Correct Answer - `13*92cm` Component velocity of proton along x-axis will be `v_x=v cos 30^@=8xx10^6xxsqrt3/2` `=6*93xx10^6ms^-1`. Component velocity of proton along y-axis will be `v_y=vsin 30^@=8xx10^6xx1/2=4xx10^6ms^-1` Since the angle between velocity component `v_x` and magnetic field B is `0^@`, therefore magnetic force on proton due to component velocity `v_x` will be `F=qv_xBsin0^@=0`. Thus the proton will move uniformly along x-axis. As the component velocity of proton along y-axis, `v_y` is perpendicular to the direction of magnetic field, therefore, the magnetic force on proton `F=qv_yBsin90^@=qv_yB` will acts as centripetal force and proton will describe a circular path due to this component velocity. As the proton covers linear distance as well as describes a circular path, hence the path of the proton will be helix. Radius of the helix will be given by `qv_yB=mv_y^2//r` or `r=(mv_y)/(qB)=(1*67xx10^(-27)xx4xx10^6)/(1*6xx10^(-19)xx0*30)` `=13*92xx10^-2m=13*92cm` |
|
| 450. |
A magnetic field of `100G(1G=10^-4T)` is required which is uniform in a region of linear dimension about `10cm` and area of cross section about `10^-3m^2`. The maximum current carrying capacity of a given coil of wire is `15A` and the number of turns per unit length that can be wound round a core is at most `1000turns m^-1`. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic. |
|
Answer» Here, `B=100G=10^-2T`, `I=15A`, `n=1000m^-1` As, `B=mu_0nI` or `nI=(B)/(mu_0)=(10^-2)/(4pixx10^-7)=(10^-2)/(4xx(22//7)xx10^-7)=7955~~8000` We may have, `I=10A`, `n=800`. The core length may be `50cm` having 400 turns and area of cross-section is `5xx10^-3m^2` which is five times given value. |
|