InterviewSolution
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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.
| 601. |
Two long straight conductors are connected radially to two arbitary points A and B of a circular conductor. Calculate the magnetic field at the center of the coil. |
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Answer» Correct Answer - Zero |
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| 602. |
The field normal to the plane of a wire of `n` turns and radis `r` which carriers `i` is measured on the axis of the coil at a small distance `h` from the centre of the coil. This is smaller than the field at the centre by the fraction.A. `3/2 (h^(2))/(r^(2))`B. `2/3 (h^(2))/(r^(2))`C. `3/2 (r^(2))/(h^(2))`D. `2/3 (r^(2))/(h^(2))` |
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Answer» Correct Answer - A |
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| 603. |
Figure shows three cases: in all case the circular part has radius r and straight ones are infinitely long. For the same current the ratio of field B at centre P in the three case `B_1:B_2:B_3` is A. `(-(pi)/2):((pi)/2):((3pi)/4-1/2)`B. `(-(pi)/2+1):((pi)/2+1):((3pi)/4+1/2)`C. `-(pi)/2:(pi)/2:3(pi)/4`D. `(-(pi)/2-1):((pi)/2-1/4):((3pi)/4+1/2)` |
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Answer» Correct Answer - A |
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| 604. |
Looking downward, an electron appears moving anticlock-wise on a horizontal circle under a magnetic field. What is the direction of the magnetic field? |
| Answer» The motion of electron in anticlock wise direction causes the current in clockwise direction. The magnetic force on the electron is along the radius towards the centre of circle. It will be so if the direction of magnetic field is vertically upwards. | |
| 605. |
shows a square current carrying loop `ABCD` of side `10cm` and current `I = 10A` The magnetic moment `oversetrarrM` of the loop is .A. `(0.05)(hati-sqrt3hatk)A-m^(2)`B. `(0.05)(hatj-sqrt3hatk)A-m^(2)`C. `(0.05)(sqrt3hati+hatk)A-m^(2)`D. `(hati+hatk)A-m^(2)` |
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Answer» Correct Answer - A Areal vects will be `bot` to the plane Area vec-tors is at `60^(@)` angle with axis and `30^(@)` angle with -z axis `oversetrarrM = NI oversetrarrA` `oversetrarrM = 10 xx 10^(-2) Amp -m^(2)` `M = 0.1 Amp - m^(2)` `oversetrarr(M) = Mx hati - My hatk = Mcos 60 hati - Mcos 30^(@)hatk` `oversetrarr(M) = 0.05 (hati - sqrt3 hatk)`. |
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| 606. |
There are two infinite long parallel straight current carrying wires, A and B separated by a distance r. Figure. The current in each wire is I. The ratio of magnitude of magnetic field at point P and Q when points P and Q lie in the plane of wires is: |
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Answer» Correct Answer - `(8)` Magnetic field at P due to currents in two wires will be acting perpendicular to the plane of wires, upwards and is given by `B_P=(mu_0)/(4pi)(2I)/((r//2))+(mu_0)/(4pi)(2I)/((r//2))=(2mu_0I)/(pir)` Magnetic field at Q due to current in A is perpendicular to the plane of wire upwards and due to current in B is perpendicular to the plane of wire downwards and is given by `B_Q=-(mu_02I)/(4pi2r)+(mu_02I)/(4pir)=(mu_0I)/(4pir)` `:. B_P/B_Q=((2mu_0I//pir))/((mu_0I//4pir))=8` |
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| 607. |
An infinite current carrying wire passes through point O and is perpendicular to the plane containing a current carrying loop ABCD as shown in figure. Choose the correct option(s). A. Net force on the loop is zero.B. Net torque on the loop is zero.C. As seen from O, the loop rotates clockwise.D. As seen from O, the loop rotates anticlockwise. |
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Answer» Correct Answer - A::C There is no force on the portion of wire AB and CD as current in them is parallel to the direction of magnetic field. The magnetic force on wire BC is perpendicular to the loop directed outwards and on wire OA, the magnetic force would be perpendicular to the loop directed inwards. Both these forces will be equal in magnitude. Therefore, the net force on the wire loop is zero and torque on the loop would be along the clockwise direction as seen from O. |
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| 608. |
Why are 2, 50 and 500 turns provided in the coil of tangent galvanometer? |
| Answer» The different turns (n) in the coil of tangent galvanometer are chosen according to the strength of current (I) to be measured. For low current, the larger number of turns in the coil of tangent galvanometer are to be used and vice-versa. This is because in tangent galvanometer, `Ixxn prop tan theta`. Due to it, the deflection `theta` will be controlled and it will be within the maximum accuracy limit. | |
| 609. |
A galvanometer coil is replaced by another coil of diameter one-fourth of the original diameter and the total number of turns as ten times the original number. What will be the new deflection if the same current is passed through it? |
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Answer» Correct Answer - `5//8` times the original deflection Case (i), `A_1=piD^2//4, n_1=n, theta_1=theta` Case (ii) `A_2=pi(D//4)^(2//4)=A_1/16`, `n_2=10n, theta=?` `I=(k)/(n_1BA_1)theta_1=(ktheta_2)/(n_2BA_2)`, so `theta_2=theta_1xx(n_2A_2)/(n_1A_2)=thetaxx10xx1/16=(5theta)/(8)` |
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| 610. |
The coil of moving coil galvanometer has an effective area `6xx10^-2m^2`. It is suspended in a magnetic field of `3xx10^-2Wbm^-2`. If the torsional constant of the suspension fibre is `5xx10^-9Nmdeg^-1`, finds its current sensitivity in degree per micro-ampere. |
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Answer» Correct Answer - `0*36 deg//mu_A` Here, `n=1`, `A=6xx10^-2m^2`, `B=3xx10^-2Wbm^-2`, `k=5xx10^-9Nmdeg^-1` Current sensitivity `I_s=theta/I=(nBA)/(k)` `=(1xx(3xx10^-2)xx(6xx10^-2))/(5xx10^-9)` `=3*6xx10^5degA^-1` `=(3*6xx10^5)/(10^6)deg//muA` `=0*36deg//muA` |
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| 611. |
A microameter has a resistance of `100 omega` and a full scale range of ` 50 muA`. It can be used as a voltmeter or as a higher range ammeter provides a resistance is added to it . Pick the correct range and resistance combination(s)A. `50V` range with `10kOmega` resistance in seriesB. `10V` range with `200kOmega` resistance in seriesC. `5mA` range with `1Omega` resistance in parallelD. `10mA` range with `1Omega` resistance in parallel |
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Answer» Correct Answer - B::C Here, `G=100Omega`, `I_g=50xx10^-6A=5xx10^-5A` For a voltmeter of range `50V`, `R=(50)/(5xx10^-5)-100=10^6-10^2=9*999xx10^5Omega` For a voltmeter of range `10V`, `R=(10)/(5xx10^-5)-100=199,900Omega` `~~200kOmega` in series For an ammeter of range `5mA`, `S=(5xx10^-5xx100)/(5xx10^-3-5xx10^-5)~~1Omega` in parallel For an ammeter of range `10mA` `S=(5xx10^-5xx100)/(10xx10^-3-5xx10^-5)=0*5Omega` in parallel |
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| 612. |
What is an ammeter? How is it used in an electric circuit? How does it differ from a voltmeter? |
| Answer» Ammeter is a low resistance galvanometer. The resistance of ammeter is low and that of voltmeter is high. Ammeter is connected in series and voltmeter in parallel in the circuit. | |
| 613. |
Why should a voltmeter have a high resistance and a low current carrying capacity? |
| Answer» Voltmeter is a high resistance galvanometer. It is used to measure potential difference between two points of the circuit. To measure a potential difference between two points of a circuit, the voltmeter is connected in parallel to the circuit across those two points. The potential difference between those two points will not be affected if there is no change in the current flowing through the circuit between those two point, which will be so if practically no current flows through voltmeter. The same is possible if voltmeter has a high resistance and a low current carrying capacity. That is why a voltmeter should have high resistance and low current capacity. | |
| 614. |
Explain giving reasons, the basic difference in converting a galvanometer into (i) an ammeter and (ii) a votmeter. |
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Answer» (i) Ammeter is a low resistance galvanometer. When a low resistance shunt is connected across galvanometer, it becomes ammeter. (ii) Voltmeter is a high resistance galvanometer. When a suitable high resistance is connected in series with the galvanometer, it becomes voltmeter. |
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| 615. |
A proton is to circulate the earth along the equator with a speed of `1*0xx10^7ms^-1`. Find the minimum magnetic field which should be created at the equator for this purpose. The mass of proton `=1*7xx10^(-27)kg` and radius of earth `=6*37xx10^6m`. |
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Answer» Correct Answer - `1*67xx10^-8T` `qv B=mv^2//r` or `B=(mv)/(qr)=(1*7xx10^(-27)xx(1*0xx10^7))/((1*6xx10^(-19))xx(6*37xx10^6))` `=1*67xx10^-8T` |
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| 616. |
Give two characteristics of a material used for making permanent magnets. |
| Answer» The material used for making permanent magnets must have high retentivity and high coercivity. | |
| 617. |
How does the intensity of magnetization of a paramagnetic material vary with increasing applied magnetic field? |
| Answer» A low field values, intensity of magnetisation increases almost linearly with field. At suffiently high field values, intensity of magnetisation gets saturated becoming independent of field. | |
| 618. |
A magnetic needle has magnetic moment of `6*7xx10^-2Am^2` and moment of inertial of `7*5xx10^-6kgm^2`. It perform `10` complete oscillations in `6*70s`. What is the magnitude of the magnetic field? |
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Answer» Here, `M=6*7xx10^-2Am^2` `I=7*5xx10^-6kgm^2` Time for one oscillation, `T=(6*70)/(10)=0*67s, B=?` From `T^**=2pisqrt((I)/(MB))`, `B=(4pi^2I)/(MT^2)=(4xx(22//7)^2xx7*5xx10^-6)/(6*7xx10^-2(0*67)^2)` `=0*01T` |
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| 619. |
In hydrozen atom, the electron is making `6.6xx10^(15)` revolution per second in a circular path of radius `0.53A^@`. What is the magnetic induction produced at the centre of the orbit? |
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Answer» Here, `v=6.6xx10^(15)rps`, `r=0.53xx10^(-10)m`, Current due to orbital motion of electron is: `I=e/T=ev=(1.6xx10^(-19))xx(6.6xx10^(15))=1.056xx10^-3A` `B=(mu_0)/(4pi)(2pinI)/(r)=(10^-7xx2xx(22//7)xx1xx(1.056xx10^-3))/((0.53xx10^(-10)))=12.5 Wb//m^2` |
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| 620. |
An `alpha` particle travels in a circular path of radius `0.45 m` in a magnetic field `B = 1.2 Wb//m^(2)` with a speed of `2.6xx10^(7) m//sec`. The `alpha` particle isA. `1.1xx10^(-5)sec`B. `1.1xx10^(-6)sec`C. `1.1xx10^(-7)sec`D. `1.1xx10^(-8)sec` |
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Answer» Correct Answer - C |
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| 621. |
A particle of charge `-16xx10^(-18) coulomb` moving with velocity `10 ms^(-1)` along the ` x- axis `, and an electric field of magnitude `(10^(4))//(m)` is along the negative ` z- ais`. If the charged particle continues moving along the ` x`- axis , the magnitude of `B` isA. `10^(3)Wb//m^(2)`B. `10^(3)Wb//m^(2)`C. `10^(16)Wb//m^(2)`D. `10^(-3) Wb//m^(2)` |
| Answer» Correct Answer - 2 | |
| 622. |
A particle of mass `M` and charge `Q` moving with velocity `vec(v)` describe a circular path of radius `R` when subjected to a uniform transverse magnetic field of induction `B`. The work done by the field when the particle completes one full circle isA. `BQv 2piR`B. `((Mv^(2))/R)2piR`C. ZeroD. `BQ2piR` |
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Answer» Correct Answer - C |
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| 623. |
A particle of mass `M` and charge `Q` moving with velocity `vec(v)` describe a circular path of radius `R` when subjected to a uniform transverse magnetic field of induction `B`. The work done by the field when the particle completes one full circle isA. `((Mv^(2))/(r))2piR`B. zeroC. `BQ,2piR`D. `BQv 2piR` |
| Answer» Correct Answer - 4 | |
| 624. |
A coil one turn is made of a wire of certain lenghth and then from the same length a coil of two turns is made. If the same current is passed both the cases, then the ratio of magnetic induction at there centres will beA. `2 : 1`B. `1 : 4`C. `4 : 1`D. `1 : 2` |
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Answer» Correct Answer - b |
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