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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A charged particle enters an environment of a strong and non-unifrom magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment ? |
| Answer» Yes, the final speed be equal to its initial speed as the magnetic force acting on the charged particle only changes the direction of velocity of charged particle but cannot change the magnitude of velocity of charged particle. | |
| 2. |
An electron emitted by a heated cathode and accelerted through a potential difference of `2.Kv,` enters a region of uniform magnetic field of `0.15T`. Determine the trajectory of the electron if the field makes an angle `30^(@)` with the initial velocity. |
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Answer» When magnetic field makes an anlge `30^(@)` with the initial velocity i.e. `theta=30^(@)` Then `V^(1)=V sin theta =sqrt((2ev)/(m))=sin 30^(@)` `V^(1)=(8)/(3)xx10^(7)xx(1)/(2)=(4)/(3)xx10^(7) m//s` The radius of the helical path is `r=(mV^(1))/(Be)` `=(9xx10^(-31)xx((4)/(3)xx10^(7)))/(0.15xx1.6xx10^(-19))=0.5 mm` |
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| 3. |
A galvanometer coil has a resistance of `12 Omega` and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V? |
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Answer» Here `G=12 Omega, Ig=3mA=3xx10^(-3) A, V=18 V` `R=(V)/(I_(g))-G=(18)/(3xx10^(-3))-12=(6000-12)Omega =5988 Omega.` |
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| 4. |
In the circuit (Fig.) the current is to be measured. What is the value of the current if the ammeter shown (a) is a galvanometer with a resistance `R_(G)=60.00 Omega,` (b) is a galvanometer described in (a) but converted to an ammeter by a shunt resistance `r_(s)=0.02 Omega,` (c) is an ideal ammeter with zero resistance ? |
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Answer» a) Total resistance in the circuit is, `R_(G)+3=63 Omega.` Hence I=3/63=0.048 A. b) Resistance of the galvanometer converted to an ammeter is, `(R_(G) r_(s))/(R_(G)+r_(s))=(60 Omega xx0.02 Omega)/((60+0.02))=0.02 Omega` Total resistance in the circuit is, `0.02 Omega + 3 Omega=3.02 Omega.` Hence I=3/3.02=0.99 A. c) For the ideal ammeter with zero resistance I=3/3=1.00 A |
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| 5. |
Two moving coil meters, `M_(1) and M_(2)` have the following particulars : `R_(1)=10 Omega, n_(1)=30`, `A_(1)=3.6xx10^(-3)m^(2), B_(1)=0.25 T` `R_(2)=14 Omega, n_(2)=42` `A_(2)=1.8xx10^(-3)m^(2), B_(2)=0.50 T` (The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of `M_(2) and M_(1)`. |
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Answer» Given, `R_(1)=10 Omega, N_(1)=30, A_(1)=3.6xx10^(-3)m^(2), B_(1)=0.25 T` `R_(2)=14 Omega, n_(2)=42, A_(2)=1.8xx10^(-3)m^(-2), B_(2)=0.50 T, K_(1)=K_(2)` a) `I=(NAB)/(K)` `(I_(s2))/(I_(s1))=(N_(2)B_(2)A_(2)K_(1))/(N_(1)B_(1)A_(1)K_(2))=(42xx0.50xx1.8xx10^(-3))/(30xx0.25xx3.6xx10^(-3))=1.4` b) `V=(NAB)/(KR)` `(V_(s2))/(V_(s1))=(N_(2)B_(2)A_(2). K_(1)R_(1))/(N_(1)B_(1)A_(1).K_(2)R_(2))=(42xx0.50xx1.8xx10^(-3)xxKxx10)/(Kxx14xx30xx0.25xx3.6xx10^(-3))=1` |
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| 6. |
A current-carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns around it self (i.e. turns about the vertical axis). |
| Answer» No, because that would require `tau` to tbe in the vertical direction. But `tau=I A xx B`, and since A of the horizontal loop is in the vertical direction, `tau` would be in the plane of the loop for any B. | |
| 7. |
A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field outside the toroid, |
| Answer» For outside the toroid, the magnetic field is zero, because the magnetic field due to toroid is only inside it and along the length of toroid. | |
| 8. |
A straight wire carrying a current of `12A` is bent into a semicircular arc of radius `2*0cm` as shown in figure. Consider the magnetic field `vecB` at the centre of arc. (a) What is the magnetic field due to the staight segments? (b) In what way the contribution to `vecB` from the semicircle differs from that of a circular loop and in what way does it resemble? (c) Would your answer be different if the wire were bent into a semicircle arc of the same radius but in the opposite way as shown in figure |
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Answer» a) dl and r for each element of the straight segments are parallel. Therefore, `dl xx r =0` Straight segments do not contribute to |B|. b) For all segments of the semicircular are `dlxxr` are all parallel to each other (into the plane of the paper). All such contributions add up in magnitude. Hence direction of B for a semicircular are is given by the right-hand ruele and magnitude is half that of a circular loop. Thus B is `1.9xx10^(-4) T` normal to the plane of the paper going into it. (c) Same magnitude of B but opposite in direction to that in (b). |
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| 9. |
If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis (Fig.), which way would the Lorentz force be for (a) an electron (negative charge), (b) a proton (positive charge). |
| Answer» The velocity v of particle is along the x-axis, while B, the magnetic field is along the y-axis, so `vxxB` is along the z-axis (screw ruel or right-hand thumb rule). So, (a) for electron it will be along-z axis. (b) for a positive charge (proton) the force is along +z axis. | |
| 10. |
A solenoid 60 cm long and of radius 7.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis, both the wire and the ais of the solenoid are in the hoorizontal plane. The wire is connected through two lead parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire ? `g=9.8 m s^(-2)`. |
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Answer» `B=mu_(0)n I` Force acts on the wire normal to its length `therefore F= BI^(1)l` `=mu_(0)n II^(1)l and n=((900)/(60))/(100)=1500` `=mu_(0)n II^(1)l = mg` `I=(2.5xx9.8)/(1000xx4pixx10^(-7)xx1500xx6xx(1)/(15))=108 A.` |
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| 11. |
What is the force on a conductor of length L carrying a current 'I' placed in a magnetic field of induction B ? When does it become maximum ? |
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Answer» i) Force on a conductor (F) `=B I L sin theta` ii) If `theta = 90^(@), F_("Max")= B I L` i.e., the direction of current and magnetic field are perpendicular to each other, then force is maximum. |
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| 12. |
What is the relation between the permittivity of free space `e_(0)`, the permeability of free space `m_(0)` and the speed of light in vaccum ? |
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Answer» Speed of light in vacuum `(C )=(1)/(sqrt(mu_(0)epsi_(0))` Here `mu_(0)=m_(0)=` permeability in vacuum `epsi_(0)`= permittivity in vacuum. |
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| 13. |
A ciruclar coil of 20 turns and radius `10cm` is placed in a uniform magnetic field of `0.1T` normal to the plane of the coil . If the current in the coil is `5.0A` what is the average force on each electron in the coil due to the magnetic field `(` The coil is made of copper wire of cross`-` sectional area `10^(-5)m^(2)` and the free electron density in copper is given to be about `10^(29)m^(-3))`. |
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Answer» n=20, r=10 cm, B=0.10 T, I=5.0 A Force on each electron n=No. of electrons per unit volume A= Area of cross-section of wire `F=Be V=(BI)/(nA)=(0.1xx5)/(10^(29)xx10^(-5))=5xx10^(-25)N` |
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| 14. |
Answer the following questions: (a) A magnetic field that varies in magnitude from point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travles undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle? (b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment? (c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path. |
| Answer» The magnetic field is in constant direction from east to west. According to the question, a charged particle travels undeflected along a straight path with constant speed. It is only possible, if the magnetic force experienced by the charged particle is zero. The magnitude of magnetic force on a moving charged particle in a magnetic field is given by `F=qvB sin theta`. Here F=0, if and only if `sin theta`. This indicates the angle between the velocity and magnetic fied is `0^(@)" or "180^(@)`. Thus, the charged particle moves parallel or antiparallel to the magnetic field B. | |
| 15. |
A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid ? |
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Answer» The number of turns per unit length is `n=(500)/(0.5)=1000` turns/ m The length l=0.5m and radius r=0.01 m Thus, `l//a=50" i.e.,"l gt gt a`. Hence, we can use the long solenoid formula, namely, Eq. `(B= mu_(0) nI) B=mu_(0)nI` `=4pi xx 10^(-7)xx10^(3)xx5=6.28xx10^(-3) T` |
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| 16. |
A circular wire loop of radius 30 cm carries a current of 3.5 A. Find the magnetic field at a point on its axis 40 cm away from the centre. |
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Answer» Radius `(r )=30 cm =30 xx10^(-2)m` Current `(i) =3.5A` `x=40 cm =40 xx 10^(-2)m` `B=(mu_(0) N i r^(2))/(2(R^(2)+x^(2))^(3//2))` `B=(4pixx10^(-7)xx1xx3.5xx(30xx10^(-2))^(2))/(2[(30xx10^(-2))^(2)+(40xx10^(-2))^(2)]^(3//2))` `=(4xx3.14xx3.5xx9xx10^(-9))/(2xx10^(-6)[900+1600]^(3//2))` `=(4xx3.14xx3.5xx9xx10^(-9))/(2xx10^(-3)[125])` `=(4xx3.14xx3.5xx9xx10^(-9))/(250)` `B=1.582xx10^(-6) T.` |
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