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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. |
Two concentric coils of 10 turns each are placed in the same plane. Their radii are 20 cm and 40 cm and carry 0.2 A and 0.3 A current respectively in opposite directions. The magnetic induction (in tesla) at the centre isA. `3/4 mu_(0)`B. `5/4 mu_(0)`C. `7/4 mu_(0)`D. `9/4 mu_(0)` |
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Answer» Correct Answer - b Two coils carry currents in opposite directions, hence net magnetic field at centre will be difference of the two fields. i.e. `B_("net")=(mu_(0))/(4pi), 2piN[(i_(1))/(r_(1))-(i_(2))/(r_(2))]=(10mu_(0))/2[0.2/0.2-0.3/0.4]=5/4 mu_(0)` |
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| 2. |
Cyclotron is used to accelerateA. negative ionB. positive ionC. electronD. None of these |
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Answer» Correct Answer - B A cylotron is a device by which the positively charged particles like proton, deuetron etc., can be acclearated. The working of cyclotron is bassed on the fact that a positively charged particle can be accelerated to a suffciently high energy with the help of similler values of oscilating electric field by making it to cross the same electric field time and again with the use of strong magentic field. |
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| 3. |
Assertion: Cyclotron is a device which is used to accelerate the position ions. Reason: Cyclotron frequency depends upon the velocity.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - C Cyclotron is suitable for accelerating heavy charged particles such as proton. `alpha`-particle and positive ions. In a cyclotron, the positive ions cross agin and again the same alternating (ration frequency) electric filed. Cyclotron is also known as magnetic resonance accelerator. Cyclotron frequency is given by `v=1/T=(Bq)/(2pim)` . It is obvious that cyclotron frequency does not depend upon velocity of charged particle. |
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| 4. |
A wire is bent in the form of an equilateral triangle PQR of side 10 cm and carrise a current of 5.0 A. It is placed in a magnetic field B of magnitude 2.0 T directd perpendicularly to the plane of the loop. Find the forces on the three sides of the triangle. |
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Answer» Suppose the field and the current have directions as shown in the figure provided. The force on PQ is `vecF=iveclxxvecB` or `F_1=5.0Axx10cmxx2.0T=1.0N` The rule of vector product shows that force `F_1` is perpedicular to PQ and is directed towards the inside of the triangle. The forces `vecF_2` and `vecF_3` QR and RP can also be obtained similarly. Both the forces are `1.0N` directed perpendicularly to the respective sides and directed inside the triangle. The three forces `vecF_1, vecF_2` and `vecF_3` will have zero resultant, so that there is no net magnetic force on the triangle. This result can be generalised. Any closed current loop, placed in a homogenous magnetic field, does not experience a net magnetic force. |
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| 5. |
An electron with kinetic energy 2.5 keV moving along the positive direction of an x axis enters a region in which a uniform electric field of magnitude 10 kV`//`m is in the negative direction of the y-axis. A uniform magnetic field `vecB` is to be set up to keep the electron moving along the x-axis , and the direction `vecB` of is to be chosen to minimize the required magnitude of `vecB`. In unit-vector notation, what `vecB` should be set up? |
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Answer» Correct Answer - `3.4xx10^(-4)T` |
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| 6. |
An electron and a proton are moving with the same kinetic energy along the same direction . When they pass through a uniform magnetic field perpendicular to the direction of their motion , they describe circular paths of the same radius. |
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Answer» `r=sqrt(2mKE)/(qB) :. Lprop(sqrtm)/q` [for constant KE and B] Here, q is same for electron and proton Therefore, Radius of proton will be more. |
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| 7. |
A proton is moving along `Z`-axis in a magnetic field. The magnetic field is along `X`-axis. The proton will experience a force alongA. X-axisB. Y-axisC. Z-axisD. Negative Z-axis |
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Answer» Correct Answer - B |
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| 8. |
A charged particle carrying charge `q=10 muC` moves with velocity `v_1=10^6 ,s^-1` at angle `45^@` with x-axis in the xy plane and experience a force `F_1=5sqrt2 mN` along the negative z-axis. When the same particle moves with velocity `v_2=10^6 ms^-1` along the z-axis, it experiences a force `F_2` in y-direction. Find the magnetic field `vecB`.A. `(10^-3T)(hati+hatj)`B. `(2xx10^-3T)hati`C. `(10^-3T)hati`D. `(2xx10^-3T)(hati+hatj)` |
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Answer» Correct Answer - c For the first case: `vecF=qvecvxxvecB` `implies -5sqrt2xx10^-3hatk` `=10^-5xx(10^6)/(sqrt2) (hati+hatj)xx(B_xhati+B_yhatj+B_zhatk)` `=(10/(sqrt2))[B_zhati-B_zhatj+(B_y+B_x)hatk]` `implies B_z=0, B_y-B_x=-10^-3T....(i)` Similarly, for the second case: `F_2hatj=(10^-5)(10^6hatk)xx[(B_xhati+B_yhatj+B_zhatk)]` `F_2hatj=10(B_xhatj-B_yhati)` `F_2=10B_x, B_y=0....(ii)` Using eqs. (i) and (ii), we get `B_x=10^-3T` Thus, `vecB=(10^-3T) hati` Also, `F_2=10B_x=10^-2N`. |
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| 9. |
If a charged particle of charge `5 muC` and mass 5 g is moving with constant speed 5 m/s in a uniform magnetic field B on a curve `x^(2)+y^(2)=25`, where x and y are in meter. The value of magnetic field will beA. 1 teslaB. 1 kilo tesla along z-axisC. 5 kilo tesla along the x-axisD. 1 kilo tesla along any line in the x-y plane |
| Answer» Correct Answer - b | |
| 10. |
In the figure shown there are two semicircles of radii `r_(1)` and `r_(2)` in which a current `i` is flowing. The magnetic induction at the centre `O` will be A. `(mu_(0) i)/(r) (r_(1) + r_(2))`B. `(mu_(0) i)/(4) (r_(1) -r_(2))`C. `(mu_(0) i)/(r) ((r_(1) + r_(2))/(r_(1) r_(2)))`D. `(mu_(0) i)/(r) ((r_(2) - r_(1))/(r_(1) r_(2)))` |
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Answer» Correct Answer - C The magnetic induction due to each semicircular parts will be in the same direction perpendicular the paper inwards. `:. B = B_(1) + B_(2) = (mu_(0)i)/(4r_(1)) + (mu_(0) i)/(4 r_(2))` `= (mu_(0) i)/(4) ((r_(1) + r_(2))/(r_(1) r_(2))) ox` |
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| 11. |
In the figure shown there are two semicircles of radii `r_(1)` and `r_(2)` in which a current `i` is flowing. The magnetic induction at the centre `O` will be A. `(mu_(0)i)/r(r_(1)+r_(2))`B. `(mu_(0)i)/4(r_(1)-r_(2))`C. `(mu_(0)i)/4((r_(1)+r_(2))/(r_(1)r_(2)))`D. `(mu_(0)i)/4((r_(1)-r_(2))/(r_(1)r_(2)))` |
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Answer» Correct Answer - C The magnetic induction due to both semicicular parts will be in the same direction perpendicular to the paper inwards. `:. B=B_(1)+B_(2)=(mu_(0)i)/(4r_(1))+(mu_(0)i)/(4r_(2))=(mu_(0)i)/4((r_(1)+r_(2))/(r_(1)r_(2))) ox` |
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| 12. |
A long straight non-conducting string carriers a charge density of `40 muC//m`. It is pulled along its length at a speed of `300 m//sec`. What is the magnetic field at a normal distance of `5 mm` from the moving string?A. `B=4.8xx10^(-7) tesla`B. `B=3.2xx10^(-7) tesla`C. `B=2.5xx10^(-7) tesla`D. `B=5xx10^(-7) tesla` |
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Answer» Correct Answer - A The moving straight behaves like a current, `i=(40xx10^(-6)C//m)(300m//sec)` `=1.2xx10^(-2) ampere` Then `B=(mu_(0)i)/(2pir)=(2xx10^(-7)xx1.2xx10^(-2))/(0.005)` `=4.8xx10^(-7) tesla` |
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| 13. |
A solid metallic cylinder carriers a direct current. The magnetic field produce by it existsA. outside the cylinder onlyB. inside the cylinder onlyC. both inside and outside the cylinderD. neither inside and outside the cylinder |
| Answer» Correct Answer - C | |
| 14. |
In order to increase the sensitivity of a moving coil galvanometer, one should decreaseA. number of turns of the coil should be increasedB. the strength of the magnetic field should be increasedC. area of coil should be increasedD. All of the above |
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Answer» Correct Answer - d Current sensitivity of the galvanometer can be increased by all three factors. |
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| 15. |
Four wires each of length `2.0` meters area bent into four loops `P,Q,R` and `S` and then suspended into uniform magnetic field. Same current is passed in each loop. Which statement is correct? A. Couple on loop `P` will be the highestB. Couple on loop `Q` will be the highestC. Couple on loop `R` will be the highestD. Couple on loop `S` will be the highest |
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Answer» Correct Answer - D Couple of force on loop `S` will be maximum because for same perimeter the area of loop will be maximum and magnetic moment of loop `=ixxA`. So, it will also be maximum for loop `S`. |
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| 16. |
An arc of a circle of raduis `R` subtends an angle `(pi)/2` at the centre. It carriers a current `i`. The magnetic field at the centre will beA. `(mu_(0)i)/(2R)`B. `(mu_(0)i)/(8R)`C. `(mu_(0)i)/(4R)`D. `(2mu_(0) i)/(5R)` |
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Answer» Correct Answer - A `B=(mu_(0))/(4pi) (thetai)/r=(mu_(0))/(4pi)xx(pi)/2xxi/R=(mu_(0)i)/(8R)` |
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| 17. |
A current carrying circular coil is bent so as to convert it into a double loop, both the loops being concentric and are carrying current in the same direction. If B is the initial magnetic field at the centre, the final magnetic field at the centre will beA. zeroB. BC. 2BD. 4B |
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Answer» Correct Answer - d `B_("Initial")=(mu_(0)i)/(2r)` `B_("final")=(mu_(0)i)/(2(r/2))+(mu_(0)i)/(2(r/2))=B_("final")=(2mu_(0)i)/r=4B_("initial")` |
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| 18. |
Magnetic field at the centre of coil of `n` turns, bent in the form of a square of side `2l`, carrying current `i`, isA. `0.5xx10^(-5) T`B. `1.25xx10^(-4) T`C. `3.0xx10^(-5) T`D. `4xx10^(-5) T` |
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Answer» Correct Answer - B `B=(mu_(0)Ni)/(2r) =(4pixx10^(-7)xx50xx2)/(2xx0.5) =1.25xx10^(-4)T` |
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| 19. |
Figure 1.31 shows the trace of the path of a charged particle in a bubble chamber. Assume that the magnetic field is into the plane of the paper with magnitude 0.4 T. The smooth spiral path occurs because the particle loses energy in ionizing molecules along the path. a. Which part of the path corresponds to higher kinetic energy for the particle? Inner or outer part of spiral. b. Is the charge positive or negative? c. The radius of curvature ranges from 70 mm to 10 mm. What is the range of values of the magnitude of momentum if the magnitude of the charge is e? |
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Answer» (a) When the particle begins its motion, the outer part of the spiral corresponds to high kinetic energy. (b) Charge is positive. (c) `p_1=qBr_1=exx0.4xx70xx10^-3=exx28xx10^-3` `p_2=qBr_2=exx0.4xx10xx10^-3=exx4xx10^-3` The range of momentum is `4exx10^-3 kgm//s to 28exx10^-3` `kgms^-1`. |
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| 20. |
A proton moving with a constant velocity passes through a region of space without any changing its velocity. If `E` and `B` represent the electric and magnetic fields, respectively. Then, this region of space may haveA. I,ii and iiiB. ii,iii and ivC. I,ii and ivD. all |
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Answer» Correct Answer - C Here the proton has no accedleration so `E = B = 0`. When `E = 0` but `B != 0`, but parallel to the motion of proton, there will be no force acting,n When `E != 0` and `E,B` and motion of proton `(v)` are mutually perpendicular, there may be no net force. Forces due to the `E` and `B` cancel each other. |
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| 21. |
A loosely wound helix made of stiff metal wire is mounted vertically with the lower end just touching the mercury in a dish. When the current from a batter is started in the metal wire through the mercury, the wire executes oscillatory motion with the lower end jumping out of and into the mercury. Explain.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - B The winding of helix carry currents in the same direction therefore they experience an attractive force pulling the lower end out of mercury. As a result of this, the circuit breaks, current becomes zero and hence the force of attraction vanishes and the helix comes back to its initial position, completing the circuit again. |
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| 22. |
Statement 1: A charged particle is moving in a circular path under the action of a uniform magnetic field as shown in Fig. During motion kinetic energy of charged particle is costant. Statement 2: During the motion, magnetic force acting on the particle is perpendicular to instantaneous velocity. A. If both statement 1 and statement 2 are ture, statement 2 is the correct explanation of statement 1.B. If both statement 1 and statement 2 are ture, statement 2 is not the correct explanation of statement 1.C. If statement 1 is true, statement 2 is false.D. If statement 1 is false, statement 2 is true. |
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Answer» Correct Answer - a `vecF=qvecVxxvecB implies vecFbotvecV` So, power produced by magnetic force is zero. `implies` Kinetic energy of particle will remain conserved. |
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| 23. |
Assertion: If a charged particle is moving on a circular path in a perpendicular magnetic field, the momentum of the particle is not changing. Reason: Velocity of the particle in not changing in the magnetic field.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - D When a charged partilcle is moving on a circular path in a magnetic field, the magnitude of velocity does not change but direction of velocity is changing every moment. Hence, velocity is changing, so momentum `(mvec(v))` is also changing. |
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| 24. |
Assertion: When a charging particle is fires in a magnetic field the radius of tis circular path is directely proportional to the kinetic energy of the particle. Reason: the centripetal force on the test charge `q_(0)` is `q_(0)vB` where `v` is the velocity of a partical and `B` is the magnetic fieldA. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - C If the path of the charged particle is circular then radius of circular path is directely proportional to the speed and mass of the particle as `r=(mv)/(q_(0)B)` `:.` Centripetal force `=(mv^(2))/r=q_(0) vB` |
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| 25. |
Statement 1: A charged particle is moving in a circular path under the action of a uniform magnetic field as shown in Fig. During motion kinetic energy of charged particle is costant. Statement 2: During the motion, magnetic force acting on the particle is perpendicular to instantaneous velocity. A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - A `vec(F)=qvec(V)xxvec(B)` `implies vec(F)bot vec(V)` So power produced by magnetic force is zero. `implies` Kinetic energy of particle will remain conserved |
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| 26. |
A charged particle with charge to mass ratio `((q)/(m)) = (10)^(3)/(19) Ckg^(-1)` enters a uniform magnetic field `vec(B) = 20 hat(i) + 30 hat(j) + 50 hat(k) T` at time t = 0 with velocity `vec(V) = (20 hat(i) + 50 hat(j) + 30 hat(k)) m//s`. Assume that magnetic field exists in large space. The pitch of the helical path of the motion of the particle will beA. `(pi)/(100)m`B. `(pi)/(125)m`C. `(pi)/(250)m`D. `(pi)/(215)m` |
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Answer» Correct Answer - D |
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| 27. |
A charged particle with charge to mass ratio `((q)/(m)) = (10)^(3)/(19) Ckg^(-1)` enters a uniform magnetic field `vec(B) = 20 hat(i) + 30 hat(j) + 50 hat(k) T` at time t = 0 with velocity `vec(V) = (20 hat(i) + 50 hat(j) + 30 hat(k)) m//s`. Assume that magnetic field exists in large space. The frequency (in Hz) of the revolution of the particle in cycles per second will beA. `(10^(3))/(pisqrt19)`B. `(10^(4))/(pisqrt38)`C. `(10^(4))/(pisqrt19)`D. `(10^(4))/(2pisqrt19)` |
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Answer» Correct Answer - B |
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| 28. |
Assertion: The poles of magnet cannot be separated by breaking into two pieces. Reason: The magnetic moment will be reduced to half when a magnet is broken into two equal pieces.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - B As we know every atom of a magnet acts as a dipole. So poles cannot be separated. When magnet is broke into two equal pieces. Magnetic moment of each part will be half of the original magnet. |
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| 29. |
A charged particle with charge to mass ratio `((q)/(m)) = (10)^(3)/(19) Ckg^(-1)` enters a uniform magnetic field `vec(B) = 20 hat(i) + 30 hat(j) + 50 hat(k) T` at time t = 0 with velocity `vec(V) = (20 hat(i) + 50 hat(j) + 30 hat(k)) m//s`. Assume that magnetic field exists in large space. During the further motion of the particle in the magnetic field, the angle between the magnetic field and velocity of the particleA. remains constantB. increaseC. decreaseD. may increse or decrease |
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Answer» Correct Answer - A |
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| 30. |
Statement 1: A magnetic field independent of time can change the velocity of a charged particle. Statement 2: It is not possible to change the velocity of a particle in a magnetic field as magnetic field does no work on the charged particle.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - C Velocity is a vector quantity even if direction change velocity is said to be changing, no matter speed remains same or different. |
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| 31. |
Statement 1: A magnetic field independent of time can change the velocity of a charged particle. Statement 2: It is not possible to change the velocity of a particle in a magnetic field as magnetic field does no work on the charged particle.A. If both statement 1 and statement 2 are ture, statement 2 is the correct explanation of statement 1.B. If both statement 1 and statement 2 are ture, statement 2 is not the correct explanation of statement 1.C. If statement 1 is true, statement 2 is false.D. If statement 1 is false, statement 2 is true. |
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Answer» Correct Answer - c Velocity is a vector quantity. Even if direction chages, velocity is said to be changing, no matter if the speed remains same or different. |
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| 32. |
`L` is a circular ring made of a uniform wire, currents enters and leaves the ring through straight conductors which, if produces, would have passed through the centre `C` of ring. The magnetic field at `C` (i) due to the straight conductors is zero (ii) due to the loop is zero (iii) due to the loop is proportional to `theta` (iv) due to loop is proportional to `(pi-theta)`A. (i),(ii)B. (ii),(iii)C. (i),(iii)D. all |
| Answer» Correct Answer - A | |
| 33. |
Constant current of `1` a flows along three branches of wire frame as shown. The frame is a combination of two equilateral triangles `ACD` and `CDE` of side `1 m`. It is place in uniform magnetic field `B=4 T` acting perpendicular to the plane of paper. The magnitude of magnitude force acting on the frame is A. `12 N`B. `24 N`C. `36 N`D. zero |
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Answer» Correct Answer - A Since `vecF_(AD)=vecF_(CD)=vecF_(CED)` Hence net force on frame `= 3vecF_(CD)` `=3i(vecixxvecb)=3xx1xx1xx4` `F=12 N` |
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| 34. |
In the plane mirror, the coordinates of image of a charged particle (initially at origin as shown in fig. after two and a half time periods are (initial velocity of charge particle `v_0` in the x-y plane and the plane mirror is perpendicular to the x-axis. A uniform magnetic field `Bhati` exists in the space. `P_0` is pitch of helix, `R_0` is raduis of helix.) A. `17P_0, 0,-2R_0`B. `3P_0, 0,-2R_0`C. `17.5P_0, 0,-2R_0`D. `3P_0, 0,2R_0` |
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Answer» Correct Answer - c After two and a half time periods, its x-coordinate will be `2.5P_0` (so x-coordinate of image will be `17.5P_0`). It will be at distance `2R_0` (equal to diametre) on the negative z-axis and y-coordinate will be zero. |
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| 35. |
A triangular loop of side l carries a current l. It is placed in a magnetic field B such that the plane of the loop is in the direction of B. The torque on the loop isA. ZeroB. IBlC. `sqrt(3)/2II^(2)B^(2)`D. `sqrt(3)/4IBl^(2)` |
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Answer» Correct Answer - D |
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| 36. |
The unit of electric current "ampere" is the current which when flowing through each of two parallel wires spaced `1 m` apart in vacuum and of infinite length will give rise to a force between them equal toA. `1 N//m`B. `2xx10^(-7)N//m`C. `1xx10^(-2)N//m`D. `4pixx10^(-7)N//m` |
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Answer» Correct Answer - B Force per unit length on two parallel current carrying conductor is given by `(F)/(1)=10^(-7)xx2(i_(1)i_(2))/(a)` ` implies (F)/(1)=10^(-7)xx2xx(1xx1)/(1)=2xx10^(-7)N//m` |
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| 37. |
A current carrying loop lies on a smooth horizontal plane. Then,A. it is possible to establish a uniform magnetic field in the region so that the loop starts rotating about its own axis.B. it is possible to establish a uniform magnetic field in the region so that the loop will tip over over about any of the point.C. it is not possible that loop will tip over about any of the point whatever be the direction of established magnetic field (uniform).D. both (a) and (b) are correct. |
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Answer» Correct Answer - b As the loop is placed in horizontal plane, so area vector is along vertical direction. From `vectau=I(vecAxxvecB)`, as `vecA` is in vertical direction, `vectau` would be the plane of loop only. So, option (a) is wrong because for rotation of loop about its own axis `vectau` must be along vertical direction. (b) is correct because we can produce torque in the plane of the loop and due to this the loop can tip over. |
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| 38. |
Assertion: A circular loop carrying current lies in `XY` plane with its centre at origin having a magnetic flux in negative `Z`-axis. Reason: Magnetic flux direction is independent of the direction of current in the conductor.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - C The direction of magnetic field due to current carrying conductor can be founded by applying right hand thumb rule or right hand palm rule. When electric current is passed through a circular conductor, the magnetic field lines near the centre of the conductor are almost straight lines. Magnetic flux direction is determined only by the direction of current. |
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| 39. |
A proton (mass m and charge `+e`) and an `alpha`-particle (mass `4` m and charge `+2e`) are projected with the same kinetic energy at right angles to the unifrom magnetic field. Which one of the following statements will be true?A. The alpha-particle will be bent in a circular path with a small radius that for the protonB. The radius of the path of the alpha-particle will be greater than that of the protonC. The alpha-particle and the proton will be bent in a circular path with the same radiusD. The alpha-particle and the proton will go through the field in a staight line |
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Answer» Correct Answer - C `r=sqrt(2mk)/(qB) i.e rpropsqrt(m)/(q)` Here kinetic energy K and B are same. `:. (r_p)/(r_alpha)=sqrt(m_p)/sqrt(malpha).(q_alpha)/q_p=sqrt(m_p)/sqrt(4m_p).(2q_p)/(q_p)=1` |
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| 40. |
An `(alpha)`-particle and a proton are both simultaneously projected in opposite direction into a region of constant magnetic field perpendicular to the direction of the field. After some time it is found that the velocity of the `(alpha)`-particle has changed in a direction by `45^(@)`. Then at this time, the angle between velocity vectors of `(alpha)`-particle and proton isA. `90^(@)`B. `45^(@)`C. `45^(@)+90^(@)`D. `(45^(@)+90^(@))/2` |
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Answer» Correct Answer - c `omega_(alpha) ((2e)B)/(4m) =1/2 (eB)/m` `omega_(p)=(eB)/m` `theta_(alpha)=45^(@)` `theta_(p)=90^(@)` from their original directions. |
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| 41. |
A charged particle moves in a gravity free space where an electric field of strength E and a magnetic field of induction B exist. State the following statements as True of False. a. If `E pi 0 and B pi 0,` velocity of the particle may remain constant. b. If `E = 0`, the particle cannot trace a circular path. c. If `E=0`, kinetic energy of the particle remains constant. |
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Answer» (a) True, (b) False, (c) True A uniform megnetic field B and a uniform electric field E exist perpendicular to each other and the particle moves along a direction perpendicular to both these fields , then forces exerted by these two fields may be opposite to each other. If magnitudes of these forces are equal, then the resultant force on the particle will become zero. Hence, the particle will move with constant velocity. Hence, (a) is correct. Obviously (b) is wrong. If E is equal to zero then the particle will experience a force due to magnetic field alone. But the force exerted by the magnetic field is always perpendicular to the direction of its motion. Hence, no power is associated with this forces. In other words, no work is done by the magnetic field on the particle. Therefore, KE of the particle will remain constant. Hence (c) is correct. |
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| 42. |
An electron moves in a uniform magnetic field and follows a spiral path as shown in Fig. 1.30. State the following statements as True or False. a. Angular velocity of the electron remains constant. b. Magnitude of velocity of the electron decreases continuously. c. Net force on the particle is always perpendicular to its direction of motion. d. Magnitude of net force on the electron decreases continuously. |
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Answer» (a) True, (b) True, (c) False, (d) True, Periode of revolution of a charged particle moving in a uniform magnetic field is given by `T=(2pim)/(qB)`. This period T does not depend upon speed of the particle. In this particular question, the moving particle is an electron. Hence, its mass and charge (q) both are constant. Magnetic field is also uni- form. Hence, its period of revolution remains constant. It means electron moves with a constant angular velocity. Hence (a) is correct. In previous question we have already discussed that if a charged particle experiences a resisting motion then it fol- lows a decreasing radius spiral path. In this question, electron is moving along a spiral path of decreasing radius. It means its speed is decreasing continuously. Hence (b) is correct. Since the speed of the electron is continuously decreasing. there- fore, it is experiencing a tangential retardation. It is possible only when the component of resultant force opposite to the direction of motion of electron has non-zero value. It means, net force on electron cannot be perpendicular to its direction of motion. Hence (c) is wrong. Since speed of the electron is decreasing continuously, therefore, the force exerted by the magnetic field `(F=qvB)` is also decreas- ing continuously. Hence, magnitude of net force acting on the electron is decreasing continuously. Hence (d) is correct. |
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| 43. |
An electron is revolving round a proton, producing a magnetic field of `16 weber//m^(2)` in a circular orbit of radius `1 Å. Its angular velocity will beA. `10^(17)rad//sec`B. `1//2pixx10^(12)rad//sec`C. `2pixx10^(12)rad//sec`D. `4pixx10^(12)rad//sec` |
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Answer» Correct Answer - A Magnetic field due to revoluton of electron `B=(mu_(0))/(4pi).(2pii)/(r)=(mu_(0))/(4pi).(2pi.((eomega)/(2pi)))/(r)=10^(-7)xx(eomega)/(r)` ` implies16=10^(-7)xx(1.6xx10^(-19)omega)/(1xx10^(-10))impliesomega=10^(17)rad//sec` |
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| 44. |
A particle of charge q and mass m starts moving from the origin under the action of an electric field `vec E = E_0 hat i and vec B = B_0 hat i` with a velocity `vec v = v_0 hat j`. The speed of the particle will become `2v_0` after a time.A. `t=(2mv_(0))/(qE)`B. `t=(2Bq)/(mv_(0))`C. `t=(sqrt3Bq)/(mv_(0))`D. `t=(sqrt3mv_(0))/(qE)` |
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Answer» Correct Answer - D `vecE` is parallel to `vecB` and `vecv` is perpendicular both. Therefore, path of the particle is a helix w3ith increasing pitch. Speed of particle at any time `t` is `v=sqrt(v_(x)^(2)+v_(y)^(2)+v_(z)^(2))….(1)` Here ` v_(y)^(2)+v_(z)^(2)=v_(0)^(2)` and `v=2v_(0)` Substituting the va,ues in equation (1) we get ` t-(sqrt3mv_(0))/(qE)` |
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| 45. |
A uniform current carrying ring of mass `m` and radius `R` is connected by a massless string as shown in Fig. 1.142. A uniform magnetic field `B_0` exists in the region to keep the ring in horizontal position, then the current in the ring is (l=length of string) .A. `(mg)/(piRB_0)`B. `(mg)/(RB_0)`C. `(mg)/(3piRB_0)`D. `(mgl)/(piR^2B_0)` |
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Answer» Correct Answer - a Torque due to magnetic field `tau_(mag)=MB_0=ipiR^2B_0....(i)` Torque due to weight about the point where string is connected `tau_("weight")=mgR....(ii)` If ring remains horizontal, then `tau_(mag)=tau_("weight")` `IpiR^2B_0=mgRimpliesI=(mg)/(piRB_0)`. |
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| 46. |
A charged particle of unit mass and unit charge moves with velocity `vecv=(8hati+6hatj)ms^-1` in magnetic field of `vecB=2hatkT`. Choose the correct alternative (s).A. The path of the particle may be `x^2+y^2-4x-21=0`B. The path of the particle may be `x^2+y^2=25`C. The path of the particle may be `y^2+z^2=25`D. The time period of the particle will be 3.14s |
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Answer» Correct Answer - (a,b,d) `|vecv|=sqrt(8^2+6^2)=10ms^-1` Radius of circular path `r=(mv)/(qB)=10/2=5m` As the magnetic field is in z-direction, hence the path of the charged particle will be circular of radius 5m in x-y plane, hence option (a) and (b) may be possible. Time period of circular path `T=(2pir)/v=(2pixx5)/10=pi` seconds. |
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| 47. |
A particle of charge q and mass m starts moving from the origin under the action of an electric field `vec E = E_0 hat i and vec B = B_0 hat i` with a velocity `vec v = v_0 hat j`. The speed of the particle will become `2v_0` after a time.A. `t=(2mv_0)/(qE)`B. `t=(2Bq)/(mv_0)`C. `t=(sqrt3Bq)/(mv_0)`D. `t=(sqrt2mv_0)/(qE)` |
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Answer» Correct Answer - d `vecE` is parallel to `vecB` and `vecv` is perpendicular to both. Therefore, path of the particle is a helix with increasing pitch. Speed of particle at any time t is `v=sqrt(v_x^2+v_y^2+v_z^2)....(i)` Here, `v_y^2+v_z^2=v_0^2` and `v=2v_0 implies v_x=sqrt3v_0` `v_xa_xt implies sqrt3v_0=(qE)/mt implies t=(sqrt3mv_0)/(qE)` |
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| 48. |
A charged particle of specific charge `alpha` moves with a velocity `vecv=v_0hati` in a magnetic field `vecB=(B_0)/(sqrt2)(hatj+hatk)`. Then (specific charge=charge per unit mass)A. path of the particle is a helixB. path of the particle is circleC. distance moved by the particle in time `t=pi/(B_0alpha) is (piv_0)/(B_0 alpha)`D. velocity of the particle after time `t=pi/(B_0alpha) is ((v_0)/2hatj+(v_0)/2hatj)` |
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Answer» Correct Answer - (b,c) `vecVbotvecB` Therefore, path of the particle is a circle. In magnetic field, speed of the particle remains constant. Therefore, distance moved by particle in time `t=pi/(B_0alpha) is v_0t or (piv_0)/(B_0alpha)` Magnitude of velocity is always `v_0`, hence option (d) is incorrect. |
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| 49. |
Consider a coaxial cable which consists of an inner wire of radius a surrounded by an outer shell of inner and outer radii b and c respectively. The inner wire carries an electric current `i_0` and the outer shell carries an equal current in opposite direction. Find the magnetic field at a distance x from the axis where (a) `xlta,` (b) `altxltb`, (c) `bltxltc` and (d)`xgtc`. Assume that the current density is uniform in the inner wire and also uniform in the outer shell. A. for `x lt a,B=(mu_(0)Ix)/(2pia^(2))`B. for `a lt x lt b,B=(mu_(0)I)/(2pix)`C. for `b lt x lt c, B=(mu_(0)I(C^(2)-x^(2)))/(2pix(c^(2)-b^(2))`D. all |
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Answer» Correct Answer - D `x lt a, B=(mu_(0)Ix)/(2pia^(2))` `a lt x lt b, B=(mu_(0)I)/(2pix)` `b lt x lt c, Bxx2pix=mu_(0)[1-(I(x^(2)-b^(2)))/(pi(c^(2)-b^(2)))]` `=(mu_(0)I)/(2pix) ((c^(2)-x^(2)))/((c^(2)-b^(2)))` |
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| 50. |
A current `i` ampere flows in a circular arc of wire whose radius is `R`, which subtend an angle `3pi//2` radian at its centre. The magnetic induction `B` at the centre is A. `mu_(0)I//R`B. `mu_(0)I//2R`C. `2mu_(0)I//R`D. `3mu_(0)I//8R` |
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Answer» Correct Answer - C For a loop, magnetic induction at centre, `B=(mu_(0))/(4pi)xx(2piI)/R` When loop subtends angle `theta` at centre, then `B=(mu_(0))/(4pi)xx(thetaI)/R` In the given problem, `theta=3pi//2` `:. B=(mu_(0))/(4pi)xx(3pi)/2xx 1/R=(3mu_(0)I)/(8R)` |
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