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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.
| 1. |
When a current carrying coil is situated in a uniform magnetic field with its magnetic moment antiparallel to the field `i)` Torque on it is maximum `ii)` Torque on it is minimum `iii) PE` of loop is maximum `iv)PE` of loop is minimumA. only `i` and `ii` are trueB. only `ii` and `iii` are trueC. only `iii` and `iv ` are trueD. only `I,ii` and `iii` are true |
| Answer» Correct Answer - 2 | |
| 2. |
If plane of coil and uniform magnetic field B(4T0 is same then torque on the current carrying coil is A. `l(piR^(2)) 4(1)`B. `(l(pi R^(2))4)/(2)`C. `(lpi R^(2)4)/(8)`D. Zero |
| Answer» Correct Answer - A | |
| 3. |
A moving coil type of galvanometer is based upon the principle thatA. a coil carrying current experiences a torque in magnetic field.B. a coil carrying current produces a magnetic field.C. a coil carrying current experiences impulse in a magnetic field.D. a coil carrying current experience a force in magnetic field. |
| Answer» Correct Answer - 1 | |
| 4. |
When a current carrying coil is placed in a uniform magnetic field of induction `B`, then a torque `tau` acts on it. If `I` is the current, `n` is the number of turns and `A` is the face area of the coil and the normal to the coil makes an angle `theta` with `B`, ThenA. `tau=BInA`B. `tau = B I nA sin theta`C. `tau=B InA cos theta`D. `tau=B I n A tan theta` |
| Answer» Correct Answer - 2 | |
| 5. |
A proton is accelerating on a cyclotron having oscillating frequency of 11 MHz in external magnetic field of 1 T. If the radius of its dees is 55 cm, then its kinetic energy (in MeV) is is `(m_(p)=1.67xx10^(-27)kg, e=1.6xx10^(-19)C)``A. 13.36B. 12.52C. 14.89D. 14.49 |
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Answer» Correct Answer - D Here, `upsilon_(c)="11 MHz "=11xx10^(6)Hz` B = 1 T, R = 55 cm = `55xx10^(-2)m,` `e=1.6xx10^(-19)C, m_(p)=1.67xx10^(-27)kg.` `therefore" K.E. "=(q^(2)B^(2)R^(2))/(2m)=((1.6xx10^(-19))^(2)xx(1)^(2)xx(5xx10^(-2))^(2))/(2xx1.67xx10^(-27))` `=23.19xx10^(-13)J` `=(23.19xx10^(-13))/(1.6xx10^(-19))=14.49xx10^(6)" eV = 14.49 MeV"` |
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| 6. |
A steady current `i` flows in a small square lopp of wire of side `L` in a horizontal plane. The loop is now folded about its middle such that half of it lies in a vertical plane. Let `vecmu_(1)` and `vecmu_(2)` respectively denote the magnetic moments due to the current loop before and after folding. ThenA. `vec(mu_(2))=0`B. `vec(mu_(1))` and `vec(mu_(2))` are in the same directionC. `|vec(mu_(1))|//|vec(mu_(2))|=sqrt(2)`D. `|vec(mu_(1))|//|vec(mu_(2))|=(1//sqrt(2))` |
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Answer» Correct Answer - C `|vec(mu)_(1)|=IA,|vec(mu)_(2)|=I(sqrt(2)A)` |
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| 7. |
Two positive charges `q_1 and q_2` are moving with velocities `v_1 and v_2` when they are at points A and B, respectively, as shown in Fig. The magnetic force experienced by charge `q_1 ` due to the other charge `q_2` is A. `(mu_(0)q_(1)q_(2)upsilon_(1)upsilon_(2))/(8sqrt(2)pi a^(2))`B. `(mu_(0)q_(1)q_(2)upsilon_(1)upsilon_(2))/(4sqrt(2)pi a^(2))`C. `(mu_(0)q_(1)q_(2)upsilon_(1)upsilon_(2))/(2sqrt(2)pi a^(2))`D. `(mu_(0)q_(1)q_(2)upsilon_(1)upsilon_(2))/(sqrt(2)pi a^(2))` |
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Answer» Correct Answer - A `|vec(F)_(B//A)|=|-(mu_(0))/(4pi)(q_(1)q_(2)upsilon_(1)upsilon_(2))/(2sqrt(2)a^(2))hat(i)|` `|vec(F)_(12)|=|-(mu_(0))/(4pi)(q_(1)q_(2)upsilon_(1)upsilon_(2))/(2sqrt(2)a^(2))hat(j)|` |
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| 8. |
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. `q(b+a)B//2m` |
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Answer» Correct Answer - B `qvB=(mv^(2))/(R)` where radius R is `(b-a),` So `qB=(mv)/((b-a))` So `v_(m i n)=(qB(b-a))/(m)` |
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| 9. |
A proton moving with a velocity of `2xx10^(6)ms^(-1)` describes circle of radius `R` in a magnetic field. The speed of an `alpha-` particle to describe a circle of same radius in the same magnitude field isA. `1xx10^(6)m//s`B. `2xx10^(6)m//s`C. `4xx10^(6)m//s`D. `8xx10^(6)m//s` |
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Answer» Correct Answer - 1 `(V_(1))/(V_(2))=(q_(1))/(q_(2))xx(m_(2))/(m_(1))` |
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| 10. |
An electron of energy 1800 eV describes a circular path in magnetic field of flux density 0.4 T. The radius of path is `(q = 1.6 xx 10^(-19) C, m_(e)=9.1 xx 10^(-31) kg)`A. `2.58xx10^(-4)m`B. `3.58xx10^(-4)m`C. `2.58xx10^(-3)m`D. `3.58xx10^(-4)m` |
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Answer» Correct Answer - B `E=(1)/(2)mv^(2) rArr v=sqrt((2E)/(m))` `r=(mv)/(Be)=(m)/(Be)sqrt((2E)/(m))=(sqrt(2mE))/(Be)` `r=(sqrt(2xx1800xx1.6xx10^(-19)xx9.1xx10^(-31)))/(1.6xx10^(-19)xx0.4)=3.58xx10^(-4)m` |
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| 11. |
A charged particle moves through a magnetic field perpendicular to its direction. ThenA. Kinetic energy changes but the momentum is constantB. the momentum changes but the kinetic energy is constantC. both momentum and kinetic energy of the particle are not constantD. both momentum and kinetic energy of the particle are constant |
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Answer» Correct Answer - B NOTE : When a charged particle enters a magnetic field at a direction perpendicuar to the direction of motion , the path of the motion is circular . In circular motion the direction of velocity changes aat every point ( the magnitude remains constant). Therefore , the tangential momentum will change at every point. But kinetic energy will remain constant as it is given by `(1)/(2) mv^(2) and v^(2)` is the square of the magnitude of velocity which does not change. |
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| 12. |
A current ` I` flows along the length of an infinitely long, straight , thin - walled pipe. ThenA. the magnetic field at all points inside the pipe is the same , but not zeroB. the magnetic field is zero only on the axis of the pipeC. the magnetic field is different at different points inside the pipeD. the magnetic field at any point inside the pipe is zero |
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Answer» Correct Answer - D There is no current inside the pipe. Therefore `oint vec(B). Vec(dl) = mu_(0)I` `I = 0` :. `B = 0` |
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| 13. |
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))(-hat(j)+hat(k))`B. `(1)/(sqrt(3))(-hat(j)+hat(k)+hat(i))`C. `(1)/(sqrt(3))(hat(i)+hat(j)+hat(k))`D. `(1)/(sqrt(2))(hat(i)+hat(k))` |
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Answer» Correct Answer - D `(d)` Magnetic induction at `(a,0,a)` due to loop in `xy` plane is in `+k` direction. Due to loop in `yz` plane, the magnetic field will be in `(+hat(i))` direction. Due to both the loops, the direction of `vec(B)` will be `(1)/(sqrt(2))(hat(i)+hat(k))` |
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| 14. |
In a region of space uniform electric field is present as `vec(E)=E_(0)hat(j)` and uniform magnetci field is present as `vec(B)=B_(0)hat(j)` . An electron is released from rest at origin. Which of the following best represent the path followed by electron after released. `(E_(0) & B_(0)`are positive constants ) |
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Answer» Correct Answer - D Magnetic force `F_(M)` will act in `+x` direction. And `F_(E)` will act in `-y` direction. Hence path followed by electron ( it will not be complete circular path ). Hence option is `( c)`. |
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| 15. |
Electric field strength `bar(E)=E_(0)hat(i)` and `bar(B)=B_(0)hat(i)` exists in a region. A charge is projected with a velocity `bar(v)=v_(0)hat(j)` at origin , thenA. It moves along helix with constant pitchB. It moves along circular path in `YZ` planeC. It moves along helix with increasing pitchD. It moves along helix with decreasing pitch |
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Answer» Correct Answer - 3 It moves along a circular path in `YZ` plane due to `bar(B)` and along a straight line path due to `vec(E)` . This combination is helix but with increasing pitch. |
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| 16. |
A circular current loop of magnetic moment `M` is in an arbitrary orientation in an external magnetic field `vec(B)`. The work done to rotate the loop by `30^(@)` about an axis perpendicular to its plane is `:`A. `MB`B. `sqrt(3)(MB)/(2)`C. `(MB)/(2)`D. zero |
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Answer» Correct Answer - 4 No work is down to rotate the loop about an axis perpendicular to its plane as `vec(M)` is directed along the axis. Work is down only when the plane of the loop rotates. |
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| 17. |
Find `oint vec(B).vec(dl)` over following loops (direction in which integration has to be performed is indicated by arrows) |
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Answer» (i) `mu_(0)i_(1)` (ii) `-mu_(0)i_(2)` (iii) `-2mu_(0)i` (iv) `-2mu_(0)i` |
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| 18. |
A wire PQR carrying a current l is bent as shown in the figure. It is placed in a uniform magnetic field B. What is the ratio of magnitude of the force on PQ to that on QR ? A. `sqrt2`B. `(1)/(sqrt2)`C. 1D. None of these |
| Answer» Correct Answer - C | |
| 19. |
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. ` 30 mA`B. ` 60 mA`C. `3 A`D. ` 6 A` |
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Answer» Correct Answer - C Magnetic field in solenoid `B = mu_(0) n I ` rArr `(B)/( mu_(0)) = ni` ( where n = number of turns per unit length) rArr `(B)/( mu_(0)) = (ni)/(L)` rArr ` 3 xx 10^(3) = ( 100 i)/( 10 xx 10^(-2))` ` I = 3 A` |
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| 20. |
The inner and outer radius of a toroid core are 28 cm and 29 cm respectively and around the core 3700 turns of a wire are wounded. If the current in the wire is 10 A, then the magnetic field inside the core of the toroid isA. `2.60 xx 10^(-2)T`B. `2.60 xx 10^(-3)T`C. `4.52 xx 10^(-2)T`D. `4.52 xx 10^(-3)T` |
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Answer» Correct Answer - A The number of turns per unit length for the given toroid `n=(N)/(2pir_("av"))` The average radius of toroid `r_("av")=(28+29)/(2)="28.5 cm "=28.5xx10^(-2)m` `therefore" "n=(3700)/(2xx3.14xx28.5xx10^(-2))=2067.27~~2067` Now, `B=mu_(0)nI=4pixx10^(-7)xx2067xx10` `=2259615.2xx10^(-7)T=2.60xx10^(-2)T` |
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| 21. |
Statement-1 A toroid produces uniform magnetic field. Statement-2 A toroid is a simple solenoid bent into the shape of a hoop, so it is like an endless solendoid.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-4B. Statement-1 is True, Statement-2, is True, Statement-2 is NOT a correct explanation for Statement-4C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - D | |
| 22. |
Statement-1 In the arrangement shown, the hoop carries a constant current. This hoop can remain stationary under the effect of magnetic field of the bar magnet Statement-2 When a magnetic dipole is placed in a non-uniform magnetic field. It experiences a force opposite to direction of external magnetic field at its centreA. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-6B. Statement-1 is True, Statement-2, is True, Statement-2 is NOT a correct explanation for Statement-6C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - C | |
| 23. |
The magnetic field at the centre of dotted circle in the arrangement shown is A. `(mu_(0)l)/(4R) + (mu_(0)l)/(2piR)`B. `(mu_(0)l)/(4R)`C. `(mu_(0)l)/(2piR)`D. zero |
| Answer» Correct Answer - C | |
| 24. |
In an attempt to increases the current sensitivity of a moving coil galvanometer, it is found that its resistance becomes double while the current sensitivity increases by 10%. The voltage sensitivity of the galvanometer changes byA. `40%`B. `-45%`C. `55%`D. `-55%` |
| Answer» Correct Answer - B | |
| 25. |
The sensitivity of a tangent galvanometer increases ifA. number of turns decreasesB. number of turns increasesC. field increasesD. number of turns remains same. |
| Answer» Correct Answer - 2 | |
| 26. |
In a tangent galvanometer, the circular coils is unwound and rewound to have twice the previous radius. As a result of this the reduction factor `(K)` of the tangent galvanometer ifA. unaffectedB. doubledC. quadrupledD. halved |
| Answer» Correct Answer - 3 | |
| 27. |
The electric current in a circular coil of two turns produced a magnetic induction of 0.2 T at its centre. The coil is unwound and then rewound into a circular coil of four turns. If same current flows in the coil, the magnetic induction at the centre of the coil now isA. 0.2TB. 0.4TC. 0.6TD. 0.8T |
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Answer» Correct Answer - D When there are two turns in the coil, then `l=2xx2pi r_(1) or r_(1)=(l)/(4pi)` `"then "B_(1)=(mu_(0)N_(1)I)/(2r_(1))=(mu_(0)2xxI)/(2xx(l//4pi))=(mu_(0)4piI)/(l)` When there are four turns in the coil, then `l=4xx pir_(2)" or "r_(2)=(l)/(8pi)` `"Then "B_(2)=(mu_(0)N_(2)I)/(2r_(2))=(mu_(0)xx4xxI)/(2xx(l//8pi))=(mu_(0)16pil)/(l)` `(B_(1))/(B_(2))=(4)/(16)=(1)/(4) or B_(2)=4B_(1)=4xx0.2T=0.8T` |
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| 28. |
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. `10^(-7)T`B. `2 xx 10^(-7) T`C. `sqrt5 xx 10^(-7) T`D. `5 xx 10^(-2) T` |
| Answer» Correct Answer - C | |
| 29. |
A straight conductor carriers a current alon the `z`-axis Consider the points `A(a,0,0),B(0,-a,0),C(-a,0,0)` and `D(0,a,0)` (i) All four points have magnetic fields of the same magnitude. (ii) All four points have magnetic fields of the different direction. (iii) The magnetic fields at `A` and `C` are in opposite directions (iv) The magnetic fields at `A` and `B` are mutually perpendicularA. All four points have magnetic fields of the same magnitudeB. All four points have magnetic fields in different directionsC. The magnetic fields at A and C are in opposite directionsD. The magnetic fields at A and B are mutually perpendicular |
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Answer» Correct Answer - A,B,C,D `B=(mu_(0)I)/(2pi r)` due to a long conductor |
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| 30. |
A loosly wound helix made of stiff wire is mounted vertically with the lower end just touching a dish of mercury when a current from the battery is started in the coil through the mercuryA. the wire oscillatesB. the wire continues makes contactC. the wire breaks contact just when the current is passedD. the mercury will expand by heating due to passage of current |
| Answer» Correct Answer - 1 | |
| 31. |
A galvanometer of resistance `25Omega` is connected to a battery of 2 volt along with a resistance in series. When the value of this resistance is `3000Omega`, a full scale deflection of 30 units is obtained in the galvanometer. In order to reduce this deflection to 20 units, the resistance in series will beA. `4513ohm`B. `5413ohm`C. `2000ohm`D. `6000 ohm` |
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Answer» Correct Answer - 1 `R=(V)/(i_(g))-G` |
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| 32. |
A solenoid 1.5 metre and 4.0 cm in diameter possesses 10 turns/cm. A current of 5.0 A is flowing throught it. Calculate the magnetic induction (i) inside are (ii) At one end on the axis of solenoid respectivelyA. `2pi xx 10^(-3)T, pi xx 10^(-3)T`B. `pi xx 10^(-3) T, 2pi xx 10^(-3)T`C. `2pi xx 10^(-3) T, 2pi xx 10&(-3) T`D. `pi xx 10^(-3)T, pi xx 10^(-3) T` |
| Answer» Correct Answer - A | |
| 33. |
Two parallel wires carrying equal currents `i_(1)` and `i_(2)` with `i_(1)gti_(2)`. When the current are in the same direction, the `10 mT`. If the direction of `i_(2)` is reversed, the field becomes `30 mT`. The ratio `i_(1)//i_(2)` isA. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - 3 | |
| 34. |
Figure shown plane figure made of a conductor located in a magnetic field along the inward normal to the plane of the figure. The magnetic field starts diminishing. Then the induced current A. at point P is clockwiseB. at point Q is anticlockwiseC. at point Q is clockwiseD. at point R is zero |
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Answer» Correct Answer - A Conseptual |
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| 35. |
A loop, carring a current i, lying in the plane of the paper, is in the field of a long straight wire with current `i_0` (inward) as shown in Fig. Find the torque acting on the loop. |
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Answer» Correct Answer - `(epsilon_(0)lB)/(sigma)` The net electric field `E=vec(E)_(1)+vec(E)_(2)` `impliesE=(sigma)/(2 epsilon_(0))+(sigma)/(2 epsilon_(0))+(sigma)/(epsilon_(0))` The net force acting on the electron is zero because it move with constant velocity `implies vec(F)_(n e t)=vec(F)_(c)_vec(F)_(m)=0` `implies|vec(F)_(c)|=|vec(F)_(m)|implieseE=evBimpliesv=(E)/(B)=(sigma)/(epsilon_(0)B)` `:.` The time of motion in side of the capacitor `=t=(l)/(v)=(epsilon_(0)lB)/(sigma)` |
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| 36. |
A straight wire carring current I is turned into a circular loop. If the magnitude of magnetic moment associated with it in M.K.S. unit is M, the length of wire will beA. `(4pi)/(M)`B. `sqrt((4pi M)/(i))`C. `sqrt((4pi i)/(M))`D. `(Mpi)/(4i)` |
| Answer» Correct Answer - B | |
| 37. |
The distance between the wires of electric mains is `12 cm`. These wires experience `4mgwt` per unit length. The value of current flowing in each wire will be if they carry current in same directionA. `4.85A`B. zeroC. `4.85xx10^(-2)A`D. `8.5xx10^(-4)A` |
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Answer» Correct Answer - 1 `(F)/(l)=(mu_(0)i^(2))/(2pid)` |
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| 38. |
An ionized gas contains both positive and negative ions . If it is subjected simultaneously to an electric field along the ` +x` - direction and a magnetic field along the ` +y` - direction and the negative ions towardws `-y` - directionA. positive ions deflect towards ` +y-direction` and negative ions towards `- y direction`B. all ions deflect towards ` +y -direction`C. all ions deflect towards ` - y -direction`D. positive ions deflect towards ` - y-direction` and negative ions towards `+ y direction` |
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Answer» Correct Answer - C Case of positively charged particle : Two forces are acting on the positively charged particle (a) due to electric field in the positive ` x- direction` (b) Force due to magnetic field. ` vec(F) = q (v hat(i)xxB hat(k)) rArr vec(F) = qvB hat(-j)` this forces will move the positively charged particle towards ` Y - axis `. Case of negatively charged particle. Two forces are acting on the negatively charged particle (a) due to eletric field in the negative ` x- direction` . (b) due to magnetic field ` vec(F) = -q(vec(v)xxvec(B))` vec(F) = -q[ v( hat (-i) xxB hat (k))]` ` vec (F) = -qvB[ hat(i) xx hat(k)] , vec(F) = qvB hat(-j)` Same direction as that of positive charge . (c) is the correct answer. |
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| 39. |
A current carrying loop in a uniform magnetic field will experienceA. force onlyB. torque onlyC. both torque and forceD. neither torque nor force |
| Answer» Correct Answer - 2 | |
| 40. |
The magnet field lines due to a bar magnet are correctly shown inA. B. C. D. |
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Answer» Correct Answer - D NOTE : magnetic lines of force form closed loops . Inside a magnet , these are directed from south to north pole . |
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| 41. |
A current carrying loop is placed in a uniform magnetic field in four different orientations , I,ii,iii & iv arrange them in the decreasing order of potential Energy` A. `Igt IIIgtIIgtIV`B. `Igt IIgt IIIgtIV`C. `Igt IVgtIIgtIII`D. `IIIgtIVgtIgtII` |
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Answer» Correct Answer - A ` U = - vec(-M). Vec(B) = -MB cos theta ` In case I , ` theta = [email protected] , U = + MB` In case II , ` theta = [email protected], U = 0 ` IN case III , ` theta = actute , U = +ve( less than + MB)` In case IV , ` theta = obtuse , U = - ve` :. IgtIIIgtIIgtIV`. |
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| 42. |
An ionized gas contains both positive and negative ions . If it is subjected simultaneously to an electric field along the ` +x` - direction and a magnetic field along the ` +y` - direction and the negative ions towardws `-y` - directionA. positive ions defect towards `+y` directionand negative ions towards `-y` directionB. all ions deflect towards `+y` directionC. all ions deflect towards `-y` directionD. positive ions deflect towards `-y` direction and negative ions towards `+y` direction. |
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Answer» Correct Answer - C If ion is positive then, it will acquire velocity in `(+ve)` x direction and hence magnetic force will act in `-y` direction `F_(M)=q(vec(v)xxvec(B))` for `(-)` ion. It will acquire velocity in `-x` direction due to electric field and hence magnetic force will act ion `-y` direction because charge is negative . Hence in both the case ions will deflect in `(-ve)` y direction. |
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| 43. |
A metallic block carrying current `I` is subjected to a uniform magnetic induction `vec(B) as shown in Figure . The moving charges experience a force ` vec(F) given by ….. Which results in the lowering of the potential of the face ……. Assume the speed of the carries to be `v` . |
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Answer» Correct Answer - A::B::C::D ` vec (F) = q(vec(v)xxvec(B)) = (-e) (-v hat (i)xx B hat(j)) = ev Bhat(k)` |
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| 44. |
For a positively charged particle moving in a `x-y` plane initially along the `x-axis` , there is a sudden change in its path due to the presence of electric and//or magnetic fields beyond `p` . The curved path is shown in the ` x- y `plane and is found to be non - circular. Which one of the following combinations is possible ? A. ` vec(E) = 0 , vec(B) = b hat(i) +chat(k)`B. ` vec(E) = a hat(i) , vec(B) = c hat(k) +a hat(i)`C. ` vec(E) = 0 , vec(B) = c hat(j) + b hat(k)`D. ` vec(E) = a hat (i) , vec(B) = bc hat(k) +b hat(j)` |
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Answer» Correct Answer - B The velocity at `P` is in the ` x- directiomn` ( given) . Let vec(v) = k hat(i)`. After `p`, the positively charged particle gets deflected in the ` x- y ` plane toward - y direction and the path is non - circular . Now , ` vec(F) = q (vec(v)xx vec(B)` rArr vec(F) = q [ k hat (i)xx(c hat (k) + a hat (i))]` for option (b) ` = q [kc hat(i) xx hat(k) + hat(k) a hat(i) xx hat (i)] = kcq hat(-j)` since in option (b) , electric field is also present ` vec (E) - a hat (i) , therfore it will also exert a force in the ` + X direction` . The net result of the two forces will be a non - circular path . Only option (b) fits for the above logic . For other option , we get some other results. |
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| 45. |
Statement-1 Path of a charged particle in a uniform and steady magnetic field cannot be parabolic Statement-2 Magnetic field cannot accelerate a charged particleA. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-2B. Statement-1 is True, Statement-2, is True, Statement-2 is NOT a correct explanation for Statement-2C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - C | |
| 46. |
In the question number 73, the given 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 5T in the horizontal direction exists such that initially the axis of the coil is in direction of the field. The coil rotates through an angle of `60^(@)` under the influence of magnetic field. The magnitude of torque on the coil in the final position isA. 25 N mB. `25 sqrt3 N m`C. 40 N mD. `40 sqrt3 N m` |
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Answer» Correct Answer - B Torque `|vec(tau)|=|vecm xx vecB|=m B sin theta` Here, m =`"10 A m"^(2), B = 5 T` Now initially `theta=0^(@)` Thus, initial torquie, `tau_(i)=0` In final position `theta=60^(@)` `therefore" "tau_(f)=m B sin 60^(@)=10xx5xx(sqrt3)/(2)=25sqrt3"N m"` |
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| 47. |
Currents of `10A, 2A` are passed through two parallel wires `A` and `B` respectively in opposite directions. If the wire `A` is infinitely long and the length of the wire `B` is 2 metre, the force on the conductor `B`, which is situated at `10cm` distance from `A` will beA. `8xx10^(-5)`newtonB. `5xx10^(-5)` netwonC. `8 pixx10^(-7)` newtonD. `4pixx10^(-7)` newton |
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Answer» Correct Answer - 1 `F=(mu_(0))/(2pi)(i_(1)i_(2))/(r)xxl` |
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| 48. |
A conductor of length 2 m carrying current 2 A is held parallel to an infinitely long conductor carrying current of 12 A at a distance of 100 mm, the force on small conductor isA. `8.6 xx 10^(-5)N`B. `6.6 xx 10^(-5)N`C. `7.6 xx 10^(-5)N`D. `9.6 xx 10^(-5)N` |
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Answer» Correct Answer - D Here, `I_(1)=2A, I_(2)=12A, r = 100mm=0.1m, l=2m` Now total force on length l of smal conductor `F=(mu_(0))/(4pi)xx(2I_(1)I_(2))/(r)l=(10^(-7)xx2xx2xx12xx2)/(0.1)=9.60xx10^(-5)N` |
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| 49. |
A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field B such that B is perpendicular to the plane of loop. What is the magnetic force acting on the loop?A. `irvec(B)`B. `2piri vec(B)`C. zeroD. `pi ri vec(B)` |
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Answer» Correct Answer - 3 The net force on curren carrying carrying loop in uniform magnetic field will be zero. |
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| 50. |
A circular loop of radius `R` is bent along a diameter and given a shapes as shown in the figure. One of the semicircles `(KNM)` lies in the ` x-z` plane with their centres and the other one `(KLM)` in the `y-z` plane with their centres at the origin. current `I` is flowing through each of the semi circles as shown in figure. (a) A particle of charge `q` is released at the origin with a velocity `vec(v) = -v_(0)hat(i)`. Find the instantaneous force `vec(F)` on the particle . Assume that space is gravity free. (b) If an external uniform magnetic field `B_(0) hat(j) ` is applied , determine the force `vec(F)_(1) and vec(F)_(2)` on the semicircles `KLM and KNM` due to the field and the net force ` vec(F)` on the loop. |
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Answer» Correct Answer - A::B::C::D (a) Magnetic field ` (vec(B))` at the origin ` = Magnetic field due to semicircle KLM + Magnetic field due to other semicircle KNM`. Therefore, ` vec(B) = (mu_(0)I)/( 4 R ) ( -hat(i) )+ (mu_(0)I)/( 4 R ) (- hat(j))` rArr ` vec(B) = - (mu_(0)I)/( 4 R) hat(i) + (mu_(0)I)/( 4 R)hat(j) = (mu_(0)I)/( 4 R)(- hat(i) + hat(j))` (b) ` vec(F)_(KLM) = vec(F)_( KNM) = BI( 2 R ) hat(i) = 2 BIRhat(i)` `vec(F)_(KM) = BI ( 2 R )hat(i) = 2 BIRhat(i)` Therefore , `vec(F)_(1) = vec(F)_(2) = 2 BIR hat(i)` or total force on the loop, ` vec(F) = vec(F)_(1) + vec(F)_(2) rArr vec(F) = 4 B I R hat(i)` |
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