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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.
| 551. |
To convert galvanometer into ammeter which one of the following is connected with the coil:A. low resistence in parallelB. high resistence in parallelC. low resistence in seriesD. high resistence in series |
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Answer» Correct Answer - a |
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| 552. |
State the rule that is used to find the direction of magnetic field acting at a point near a current carrying straight conductor. |
| Answer» The direction of magnetic field acting at a point near a current carrying straight conductor can be given by Right Hand thumb rule. According to this rule, if we imagine the linear wire conductor to be held in the grip of the right hand so that the thumb points in the direction of current, then the curvature of the fingers around the conductor will represent the direction of magnetic field lines. | |
| 553. |
A straight wire of length `0.5` metre and carrying a current of `1.2` ampere is placed in a uniform magnetic field of induction 2 tesla. If the magnetic field is perpendicular to the length of the wire , the force acting on the wire isA. 2.4 NB. 1.2 NC. 3.0 ND. 2.0 N |
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Answer» Correct Answer - b |
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| 554. |
Name the physical quantity whose unit is tesla. Hence define a tesla. |
| Answer» Tesla is the SI unit of magnetic field induction or magnetic flux density at a point in the magnetic field. The magnetic field induction at a point in a magnetic field is said to be 1 tesla if one coulomb charge while moving with a velocity of `1m//s`, perpendicular to the magnetic field experience a force of `1N` at that point. | |
| 555. |
A straight wire of length `0.5` metre and carrying a current of `1.2` ampere is placed in a uniform magnetic field of induction 2 tesla. If the magnetic field is perpendicular to the length of the wire , the force acting on the wire isA. `2.4N`B. `1.2N`C. `3.0N`D. `2.0N` |
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Answer» Correct Answer - B |
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| 556. |
Name the physical quantity whose unit is tesla. Hence define a tesla.A. magnetic fluxB. magnetic fieldC. magnetic inductionD. magnetic moment |
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Answer» Correct Answer - c |
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| 557. |
A current carrying coil is subjected to a uniform magnetic field. The coil will orient so that its plane becomeA. inclined at `45^(@)` to the magnetic fieldB. inclined at any arbitary angle to the magnetic fieldC. parallel to the magnetic fieldD. perpendicular to magnetic field |
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Answer» Correct Answer - c |
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| 558. |
Calculate the magnetic field intensity at a distance of `20cm.` from a pole of strength `40Am` in air. Find the magnetic induction at the same point. |
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Answer» Correct Answer - `79*54Am^-1`; `10^-4Wb//m^2` Here, `d=20cm=1//5m, m=40Am`, `H=?, B=?` `H=(1)/(4pi)(mxx1)/(d^2)=(1xx7)/(4xx22)xx (40)/((1//5)^2)=79*54Am^-1` `B=mu_0H=4pixx10^-7xx79*54=10^-4Wbm^-2` |
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| 559. |
The magnetic field lines of a magnet form……………….loops unlike………………field lines. |
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Answer» Correct Answer - closed continuous ; electric |
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| 560. |
Isogonic lines are the lines joining places of…………………….. . |
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Answer» Correct Answer - equal declination |
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| 561. |
On what interaction is the principle of galvanometer based? |
| Answer» The principle of galvanometer is based on the interaction of current and magnetic field. | |
| 562. |
State the principle of moving coil galvanometer. |
| Answer» The principle of working of a moving coil galvanometer is based on the fact that when a current carrying coil is placed in a uniform radial magnetic field, it expriences a torque. | |
| 563. |
State the underlying principle of working of a moving coil galvanometer. Write two reasons why a galvanometer can not be used as such to measure current in a given circuit. |
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Answer» The principle of working of a moving coil galvanometer is based on the fact that if a current carrying coil is placed in a uniform radial field, it experiences a torque. A galvanometer cannot be used as such to measures current in a circuit due to following two reasons: (i) Galvanometer is very sensitive. It gives full scale deflection with a quite small current (nearly few micro ampere). (ii) In order to measrue the current, the galvanometer has to be connected in series of circuit. As its resistance is large, its presence in the circuit will decrease the effective current in the circuit. |
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| 564. |
A length of wire carries a steady current I. It is bent first to form a circular plane coil of one turn. The same length is now bent more sharply to give double loop of smaller radius. If the same current I is passed, the ratio of the magnitude of magnetic field at the centre with its first value is:A. A quarter of its first valueB. UnalteredC. Four times of its first valueD. A half of its first value |
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Answer» Correct Answer - C |
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| 565. |
A bar magnet of length `10cm` has a pole strength of `20Am`. Find the magnetic field strength at a distance of `10cm` from its centre on (i) its axial line and on (ii) its equatorial line. |
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Answer» Correct Answer - (i) `7*1xx10^-3T` (ii) `1*43xx10^-4T` Here, `2l=10cm=0*1m, m=20Am` `d=10cm=0*1m` (i) On axial line `B_1=(mu_0)/(4pi)(2M)/((d^2-l^2)^2)=(mu_0)/(4pi)(2xx(mxx2l))/((d^2-l^2)^2)` `=(10^-7xx2xx20xx0*1)/([(0*1)^2-((0*1)/(2))^2]^2)=(4xx10^-7)/((0*0075)^2)` `=7*5xx10^-3T` (ii) On equatorial line `B_2=(mu_0)/(4pi)(M)/((d^2+l^2)^(3//2))=(10^-7xx20xx0*1)/((0*1^2+0*05^2)^(3//2))` `=1*43xx10^-4T` |
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| 566. |
A neutral point is found on the axis of a bar magnet at a distance of `10cm` from one end. If the length of the magnet is `10cm` and `H=0*3gauss`, find the magnetic moment of the magnet. |
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Answer» Correct Answer - `0*506Am^2` Here, `2l=10cm`, `H=0*3G`, `M=?` `r=10+l=10+5=15cm=15xx10^-2m` As neutral point is on the axis of bar magnet, therefore, `H=(mu_0)/(4pi)(2M)/(r^3)=10^-7(2xxM)/((15xx10^-2)^3)` `M=((15xx10^-2)^3H)/(2xx10^-7)=(3375xx10^-6xx0*3xx10^-4)/(2xx10^-7)` `M=506*25xx10^-3=0*506Am^2` |
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| 567. |
How many neutral points on a horizontal board are there when a magnet is held vertically on the board? |
| Answer» There will be only one neutral point on the horizontal board. This is because field of earth is from south to north, and the field of magnetic pole on the board is radially outwards. At any point towards south of magnetic pole, field of earth and field of magnetic pole will cancel out to give a neutral point. | |
| 568. |
A current carrying rectangular coil is placed in a uniform magnetic field. In which orientation, the coil will not tend to rotateA. The magnetic field is parallel to the plane of the coilB. The magnetic field is perpendicular to the plane of the coilC. The magnetic field is at `45^(@)` with the plane of the coilD. Always in any orientation |
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Answer» Correct Answer - B |
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| 569. |
A coil carrying electric current is placed in uniform magnetic fieldA. Torque is formedB. E.M.f. is inducedC. Both (a) and (b) are correctD. None of these |
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Answer» Correct Answer - A |
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| 570. |
A coil carrying electric current is placed in uniform magnetic fieldA. torque is formedB. emf is inducedC. Both (a) and (b) are correctD. None of the above |
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Answer» Correct Answer - a |
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| 571. |
A current loop `ABCD` is held fixed on the plane of the paper as shown in figure. The arcs `BC( radius = b) and DA ( radius = a)` of the loop are joined by two straight wires `AB and CD` at the origin `O` is 30^(@)`. Another straight thin wire with steady current `I_(1)` flowing out of the plane of the paper is kept at the origin . The magnitude of the magnetic field (B) due to the loop `ABCD` at the origin (o) is :A. `(mu_(0)I)/(4pi)[[b-a)/(ab)]`B. `(mu_(0)I)/(4pi) [2(b-a) + (pi)/(3) (a +b)]`C. zeroD. `(mu_(0)I(b-a))/(24ab)` |
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Answer» Correct Answer - 2 Magnetic filed due to loop `ABCD` `= (mu_(0)I)/(4pi)((pi)/(6)) xx [(1)/(a) - (1)/(b)] = (mu_(0)I)/(24) [(b-a)/(ab)]` . |
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| 572. |
An electron enters a region where magnetic field (B) and electric field (E ) are mutually perpendicular, thenA. it will always move in the direction of BB. it will always move in the direction of EC. it always possess circular motionD. it can go undeflected also |
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Answer» Correct Answer - d |
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| 573. |
An electron enters a region where magnetic field (B) and electric field (E ) are mutually perpendicular, thenA. It will always move in the direction of `B`B. It will always move in the direction of `E`C. It always possess circular motionD. It can go undeflected also |
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Answer» Correct Answer - D |
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| 574. |
Three rings, each having equal radius R, are placed mutually perpendicular to each other and each having its centre at the origin of coordinate system. If current I is flowing through each ring, then the magnitude of the magnetic field at the common centre is A. `sqrt3 (mu_(0)I)/(2R)`B. zeroC. `(sqrt2-1) (mu_(0)I)/(2R)`D. `(sqrt3-sqrt2) (mu_(0)I)/(2R)` |
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Answer» Correct Answer - A `oversetrarrB=B_(x) hati + B_(y) hatj + B_(z) hatk` `B =sqrt(B_(x)^(2) + B_(y)^(2) + B_(z)^(2))` `"since" B_(x) = B_(y) = B_(z) so B= sqrt3 B_(0)` `B = sqrt3 . (mu_(0))/(4pi) (2piI)/R = sqrt3 (mu_(0)I)/(2R)` . |
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| 575. |
Two concentric coil X and Y of radii `16cm` and `10cm` respectively lie in the same vertical plane containing the north-south direction. Coil X has 20 turns and carries a current of 16A, coil Y has 25 turns and carries a current of 18A. The sense of current in X is anti-clockwise and in Y, clockwise, for an observer looking at the coil facing west, Figure. Give the magnitude and direction of the net magnetic field due to the coils at their centre. |
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Answer» For coil X. `r=16cm=0*16m`, `n=20`, `I=16A` Magnetic field induction at the centre of the coil X is given by `B_x=(mu_0)/(4pi)(2pinI)/(r)=(10^-7xx2pixx20xx16)/(0*16)=4pixx10^-4T` The direction of magnetic field induction `vecB_x` is towards east. For coil Y. `I=18A`, `n=25`, `r=10cm=0*10m` Magnetic field induction at the centre of coil Y is given by `B_y=(mu_0)/(4pi)(2pinI)/(r)=(10^-7xx2pixx25xx18)/(0*10)=9pixx10^-4T` The direction of magnetic field induction `B_y` is towards west `:.` Net magnetic field `=B_y-B_x=9pixx10^-4-4pixx10^-4=5pixx10^-4=1*6xx10^-3T` (Towards West) |
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| 576. |
A current I flows around a closed path in the horizontal plane of circle as shown in the figure The path consists of eight arcs with alternating radii r and `2r` . Each segment of arc subtends equal angle at the common centre `P` The magnetic field produced by current path at point `P` is .A. `(3)/(8)(mu_(0)I)/( r )` , perpendicular to the plane of the paper and directed inward .B. `(3)/(8)(mu_(0)I)/( r )` , perpendicular to the plane of the paper and directed outward .C. `(3)/(8)(mu_(0)I)/( r )` , perpendicular to the plane of the paper and directed inward .D. `(3)/(8)(mu_(0)I)/( r )` , perpendicular to the plane of the paper and directed outward . |
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Answer» Correct Answer - A `B =(mu_(0))/(4pi) (I)/(2r) xx pi + (mu_(0))/(4pi) (I)/(r) xx pi` `=(mu_(0)I)/(4r)[(1)/(2) +1]=(3)/(8) (mu_(0I))/(r)o.` . |
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| 577. |
A horizontal rod of mass `10g` and length `10cm` is placed on a smooth plane inclined at an angle of `60^@` with the horizontal with the length of the rod parallel to the edge of the inclined plane. A uniform magnetic field induction B is applied vertically downwards. If the current through the rod is `1*73ampere`, the value of B for which the rod remains stationary on the inclined plane isA. 1.73 TeslaB. `1/1.73` TeslaC. 1 TeslaD. None of the above |
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Answer» Correct Answer - C |
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| 578. |
The maximum value of the permeability of `mu` metal (77% Ni, 16% Fe, 5% Cu, 2% Cr) is `0.128 Tm A^(-1)`. Find the maximum relative permeability and susceptibility. |
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Answer» Correct Answer - `mu_r=1*0xx10^5`; `chi=1*0xx10^5` Here, `mu=0*126TmA^-1` `:. mu_r=(mu)/(mu_0)=(0*126)/(4pixx10^-7)=1*0xx10^5` Max. susceptibility `chi=mu_r-1=1*0xx10^5-1` `=1*0xx10^5` |
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| 579. |
A toroid of n turns, mean radius R and cross-sectional radius a carries current I. It is placed on a horizontal table taken as x-y plane. Its magnetic moment `vecM`A. is non-zero and points in the z-direction by symmetryB. points along the axis of the toroid `(vecM=Mhatphi)`C. is zero, otherwise there would be a field falling as `1/r^3` at large distance outside the toroidD. is pointing radially outwards. |
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Answer» Correct Answer - C In case of a toroid, the magnetic field is only confined inside the body of toroid in the form of concentric magnetic lines of force and there is no magnetic field outside the body of toroid. Thus the magnetic moment of toroid is zero. Hence, option (c) is correct. |
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| 580. |
The dimension of `sqrt((mu)/(varepsilon)` where `mu` is permeability & `varepsilon` is permittivity is same as: .A. ResistanceB. InductanceC. CapacitanceD. None of these |
| Answer» Correct Answer - A | |
| 581. |
Which of the following is a unit of permeability ?A. Amp/metreB. Amp/`m^(2)`C. HenryD. Henry/metre |
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Answer» Correct Answer - D |
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| 582. |
A charged particle would continue to move with a constant velocity in a region wherein,A. `vecE=0, vecB!=0`B. `vecE!=0, vecB!=0`C. `vecE!=0, vecB=0`D. `vecE=0, vecB=0` |
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Answer» Correct Answer - A::B::D A charged particle will move with a constant velocity in a region if the force on it due to electron field as well as magnetic field is zero or force on charged particle due to electric field is equal and opposite to the force on it due to magnetic field. Force on charged particle due to electric field, `vecF_E=qvecE` Force on charged particle due to magnetic field, `vecF_m=q(vecvxxvecB)` or `F_m=qvBsintheta`. Now, `F_E=0` if `E=0` and `F_m=0` if `sintheta=0` or `theta=0^@` or `180^@` A charged particle would continue to move with a constant velocity in a region if the force due to electric field and magnetic field is zero, i.e., (i) `E=0` and particle is moving along the direction of magnetic field (i.e., `theta=0^@` or `180^@`). Hence, `B!=0`. Option (a) is correct. (ii) `E=0` and `B=0`. Option (d) is correct. (iii) The resultant force, `qvecE+q(vecvxxvecB)=0`. In this case `E!=0` and `B!=0`, i.e., Option (b) is correct. (iv) If `B=0` but `E!=0`, then the charged particle will be accelerated by electric field. Option (c) is wrong. |
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| 583. |
The wire shown in the figure, carries a current of `60A`. Determine the magnitude of the magnetic field induction at O. Given radius of the bent coil is `2cm`. |
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Answer» Correct Answer - `1*414xx10^-3T` Mag. Field at O, `B=int (mu_0)/(4pi)*(idlsin90^@)/(r^2)` `=(mu_0)/(4pi)(i)/(r^2)intdl=(mu_0)/(4pi)(i)/(r^2)((3pir)/(2))` |
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| 584. |
The circular loop PQRSTP formed by two circular segments of radii `R_1` and `R_2`, carries a current `I`. What is the magnetic field induction at the centre O? What will be the magnetic field induction at O if `alpha=90^@`? |
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Answer» Let `hatn` be the unit vector perpendicular to the plane of loop downwards. Total magnetic field induction at O due to current through circular loop is `vecB=vecB_(PQ)+vecB_(QR)+vecB_(RS)+vecB_(STP)` `=0+(mu_0)/(4pi)I/R_2alphahatn+0+(mu_0)/(4pi)I/R_1(2pi-alpha)hatn` `=(mu_0I)/(4pi)[alpha/R_2+((2pi-alpha))/(R_1)]hatn` or `B=(mu_0I)/(4pi)[(alpha)/(R_2)+((2pi-alpha))/(R_1)]` Directed towards If `alpha=90^@`, then `B=(mu_0I)/(4pi)[(pi//2)/(R_2)+(3pi//2)/(R_1)]` `=(mu_0I)/(8)[(1)/(R_2)+(3)/(R_1)]` |
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| 585. |
If a long horizontal conductor is bent as shown in figure and a current I is passed in it, find the magnitude and direction of magnetic field induction at the centre of circular part. |
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Answer» Let `hatn` be the unit vector perpendicular to the plane of paper upwards. Magnetic field induction at O due to current through the entire structure is `vecB=vecB_(CG)+vecB_(DEF)=(mu_0)/(4pi)(2I)/(r)hatn+(mu_0)/(4pi)(2piI)/(r)(-hatn)` `=(mu_0I)/(2r)[1/pi-1]hatn=(mu_0I)/(2r)[1/pi-1]upwards` |
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| 586. |
The current loop abcde formed by two circular segments of radii `r_1(=4cm)` and `r_2(=6cm)` with common centre at O, carries a current `I(=2A)` as shown in figure. What is the magnetic field at the common centre O? What will be the value of magnetic field at O when `theta=90^@`? |
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Answer» Magnetic field at the centre O due to current through the whole current loop is `B=B_(ab)+B_(bc)+B_(cd)+B_(dea)` directed normally downwards `=0+(mu_0)/(4pi)I/r_1theta+0+(mu_0)/(4pi)I/r_1(2pi-theta)` `=(mu_0I)/(4pi)[theta/r_2+(2pi-theta)/(r_1)]` directed normally downwards If `theta=90^@=(pi//2)rad`, then `B=(mu_0I)/(4pi)[(pi//2)/(r_2)+((2pi-pi//2))/(r_1)]=(mu_0I)/(8pi)[1/r_2+3/r_1]` `=((4pixx10^-7)xx2)/(8pi)[100/6+100/4]` `=41*7xx10^-7T` |
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| 587. |
A conducting wire of length l is turned in the form of a circular coil and a current I is passed through it. For the torque, due to magnetic field produced at its centre, to be maximum, the number of turns in the coil will beA. oneB. twoC. threeD. more than three |
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Answer» Let n be the no. of turns formed of circular coil of radius r, in length l of the wire. Then `l=(2pir)n` or `r=(l)/(2pin)` Maximum torque, `tau_(max.)=nIAB=nIpir^2B` `=nIpixx(l//2pin^2)B=(l^2IB)/(4pin)` `tau` will be maximum if n is minimum. If will be so if `n=1`. |
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| 588. |
When a coil carrying current is set with its plane perpendicular to the direction of magnetic field, then torque on the coil is……………. . |
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Answer» Correct Answer - zero |
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| 589. |
Assertion : Torque on the coil is the maximum, when coil is suspended in a radial magnetic field. Reason : The torque tends to rotate the coil on its own axis.A. If both assertion and reason are true and the reason is the correctexplanation of the assertion.B. If both assertion and reason are true but reason is not the correctexplanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - B |
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| 590. |
What is the magnitude of torque which acts on a coil carrying current placed in a uniform radial magnetic field? |
| Answer» Torque, `tau=nBIA`, where the terms have their usual meanings. | |
| 591. |
In a moving coil galvanometer, there is a coil of copper having number of insulated turns N, each of area A. The coil is suspended in a radial magnetic field B. The moment of inertia of the coil about its rotational axis is I. The scale divisions in the galvanometer are n and resistance of the coil is R. If a current `i_0` in the coil produces a deflection of `pi//3` radian to the pointer of galvanometer, the value of torsional constant of the spring isA. `(NBAi_0)/(3pi)`B. `(3NBAi_0)/(2pi)`C. `(NBAi_0pi)/(3)`D. `(3NBAi_0)/(pi)` |
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Answer» Correct Answer - D As `i=(k)/(NBA)theta`, so `k=(NBAi)/(theta)=(NBAi_0)/(pi//3)=(3NBAi_0)/(pi)` |
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| 592. |
A rectangular coil of area `5.0xx10^(-4) m^2` and 60 turns is pivoted about one of its vertical sides. The coil is in a radial horizontal field of 90 G (radial here means the field liners are in the plane of the coil for any rotation). What is the torsional constant of the hair spring connected to the coil if a current of 2.0 mA produces an angular deflection of `18^@`? |
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Answer» Here, `A=5*0xx10^-4m^2`, `n=60`, `B=90G=90xx10^-4T`, `k=?` `I=2*0mA=2*0xx10^-3A, theta=18^@`, As, `I=(k)/(nBA)theta` `:. k=(nBAI)/(theta)` `=(60xx90xx10^-4xx5*0xx10^-4xx2*0xx10^-3)/(18)` `=3*0xx10^-9Nm` per degree |
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| 593. |
An electron is travelling in east direction and a magnetic field is applied in upward direction then electron will deflect inA. SouthB. NorthC. WestD. East |
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Answer» Correct Answer - B |
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| 594. |
If `theta_1` and `theta_2` be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip `theta` is given byA. `cot^2theta=cot^2theta_1+cot^2theta_2`B. `tan^2theta=tan^2theta_1+tan^2theta_2`C. `cot^2theta=cot^2theta_1-cot^2theta_2`D. `tan^2theta=tan^2theta_1-tan^2theta_2` |
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Answer» Here, `tan theta_1=(V)/(H_1)` and `tan theta_2=(V)/(H_2)` and `tan theta=V/H` `:. H_1=V cot theta_1, H_2=Vcot theta_2, H=Vcot theta` As `H^2=H_1^2+H_2^2` `:. V^2cot^2theta=V^2cot^2theta_1+V^2cot^2theta_2` Hence `cot^2theta=cot^2theta_1+cot^2theta_2` |
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| 595. |
A horizontal wire `0*1m` long having mass `3g` carries a current of `5A`. Find the magnitude of the magnetic field which must act `30^@` to the length of the wire inorder to support its weight. |
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Answer» Correct Answer - `0*1176T` `F=Bil sin theta=mg` or `B=(mg)/(il sin theta)=((3xx10^-3)xx9*8)/(5xx0*1xxsin30^@)=0*1176T` |
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| 596. |
A horizontal wire `0*2m` long carries a current of `4A`. Find the magnitude and direction of the magnetic field, which can support the weight of the wire. Given, the mass of the wire is `3xx10^-3kg//m`, `g=10ms^-2`. |
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Answer» Here, mass of wire, `m=(3xx10^-3)xx0*2kg, I=4A, l=0*2m` In equilibrium position, force `(F=IlB)` on wire carrying current I due to magnetic field is equal to weight (`mg`) of wire, i.e., `IlB=mg` or `B=(mg)/(Il)=((3xx10^-3)xx0*2xx10)/(4xx0*2)` `=7*5xx10^-3T` The weight of wire can be supported by forces F if `vecF` is acting vertically upwards (i.e., opposite to the weight of wire). It will be so if the direction of `vecB` is horizontal and perpendicular to wire carrying current. |
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| 597. |
A horizontal wire `0*1m` long carries a current of `5A`. Find the magnitude and direction of the magnetic field, which can support the weight of the wire. Given the mass of the wire is `3xx10^-3 kg//m` and `g=10ms^-2`.A. `6xx10^-3T`, acting vertically upwardsB. `6xx10^-3T`, acting horizontally perpendicular to wireC. `6xx10^-2T`, acting vertically downwardsD. `6xx10^-2T`, acting horizontally perpendicular to wire |
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Answer» Correct Answer - B Here, mass of wire `m=0*1xx(3xx10^-3)kg`, `I=5A`, `B=?`, `l=0*1m` In equilibrium position, `F=IlB=mg` or `B=(mg)/(Il)=((0*1xx3xx10^-3)xx10)/(5xx0*1)=6xx10^-3T` The weight of wire can be supported by force F if it is acting vertically upwards (i.e. opposite to the weight of wire). It will be so if the direction of `vecB` is horizontal and perpendicular to wire carrying current. |
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| 598. |
A wire placed along north-south direction carries a current of `5A` from south to north. Find the magnetic field due to a `1cm` piece of wire at a point `200cm` north east from the piece. |
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Answer» Correct Answer - `8*8xx10^(-10)T`, acting vertically downwards `dB=(mu_0)/(4pi)(Idlsintheta)/(r^2)=(10^-7xx5xx10^-2xxsin45^@)/((2)^2)` `=8*8xx10^(-10)T` Its direction is vertically downwards. |
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| 599. |
A long straight wire carries a current of `35A`. What is the magnetic of the field `vecB` at a point `20cm` from the wire? |
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Answer» Here, `I=35A`, `r=20cm=20xx10^-2m`, `B=?` `B=(mu_0)/(4pi)(2I)/(r)=10^-7xx(2xx35)/(20xx10^-2)=3*5xx10^-5T` |
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| 600. |
A long straight wire in the horizontal plane carries a current of 50A in north to south direction. Give the magnitude and direction of `vecB` at a point `2*5m` east of the wire. |
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Answer» Here, `I=50A`, `r=2*5m`, `B=?` `B=(mu_0)/(4pi)(2i)/(r)=10^-7xx(2xx50)/(2*5)=4xx10^-6T` It acts upwards, perpendicular to the plane of wire. |
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