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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.
| 501. |
A charged particle of charge q and mass m enters perpendiculalry in a magnetic field B. Kinetic energy of particle E, then frequency of rotation isA. `(qB)/(mpi)`B. `qB/(2pim)`C. `(qBE)/(2pim)`D. `(qB)/(2piE)` |
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Answer» Correct Answer - b |
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| 502. |
A charged particle of charge q and mass m enters perpendiculalry in a magnetic field B. Kinetic energy of particle E, then frequency of rotation isA. `(Bq)/(2pim)`B. `(Bq)/(2pirm)`C. `(2pim)/(Bq)`D. `(Bm)/(2piq)` |
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Answer» Correct Answer - A |
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| 503. |
When a charged particle enters a uniform magnetic field its kinetic energyA. Remains constantB. IncreasesC. DecreasesD. Becomes zero |
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Answer» Correct Answer - A |
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| 504. |
A particle of charge `q` and mass `m` moving with a velocity `v` along the x-axis enters the region `xgt0` with uniform magnetic field `B` along the `hatk` direction. The particle will penetrate in this region in the `x`-direction upto a distance `d` equal toA. ZeroB. `(mv)/(qB)`C. `(2mv)/(qB)`D. Infinity |
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Answer» Correct Answer - B |
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| 505. |
A charged particle enters a magnetic field `H` with its initial velocity making an angle of `45^(@)` with `H`. The path of the particle will beA. A straight lineB. A circleC. An ellipseD. A helix |
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Answer» Correct Answer - D |
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| 506. |
What should be the current in a circular coil of radius `5cm` to annual `B_(H)=5xx10^(-5)T`A. `0.4A`B. `4A`C. `40A`D. `1A` |
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Answer» Correct Answer - B |
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| 507. |
A current of 2 amp. flows in a long, straight wire of radius `2 mm`. The intensity of magnetic field on the axis of the wire isA. `((mu_(0))/(pi))xx10^(3)` TeslaB. `((mu_(0))/(2pi))xx10^(3)` TeslaC. `((2mu_(0))/(pi))xx10^(3)` TeslaD. Zero |
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Answer» Correct Answer - D |
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| 508. |
Using the relation for potential energy of a current carrying planar loop in a uniform magnetic field, obtain expression for work done in moving the planar loop from its unstable equilibrium position to the stable position. |
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Answer» Potential energy of a current loop when its dipole moment `vecM` makes angle `theta` with the field `vecB` is given by `U=-MB cos theta` Work done in turning the loop from orientation `theta_1` to `theta_2` is `W=-MB(cos theta_2-cos theta_1)` In unstable equilibrium, `theta_1=180^@`, and in stable equilibrium `theta_2=0^@` `:. W=-MB(cos 0^@-cos 180^@)=-MB(1+1)` `=-2MB=-2(NIA)B` |
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| 509. |
Does the torque on planar current loop in a magnetic field change when its shape is changed without changing its geometrical area. |
| Answer» Torque on a planar current loop is given by `tau=niAB sin alpha`, which is independent of shape if area A is unchanged. Hence torque on a planar current loop in a magnetic field does not change when its shape is changed without changing its area. | |
| 510. |
A thin circular wire carrying a current `I` has a magnetic moment `M`. The shape of the wire is changed to a square and it carries the same current. It will have a magnetic momentA. `M`B. `4/(pi^(2))M`C. `4/(pi)M`D. `(pi)/4M` |
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Answer» Correct Answer - D |
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| 511. |
A wire carrying current `I` has the shape as shown in the adjoining figure. Linear parts of the wire are very long and parallel to X-axis while semicicular portion of radius `R` is lying in `Y-Z` plane. Magnetic field at point `O` is A. `B=(mu_(0))/(4pi)(I)/(R)(pihati+2hatk)`B. `B=-(mu_(0))/(4pi)(I)/(R)(pihati-2hatk)`C. `B=-(mu_(0))/(4pi)(I)/(R)(pihati+2hatk)`D. `B=(mu_(0))/(4pi)(I)/(R)(pihati-2hatk)` |
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Answer» Correct Answer - a |
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| 512. |
An electron moving in a circular orbit of radius `r` makes `n` rotation per secound. The magnetic field produced at the centre has magnitudeA. `(mu_(0)"ne")/(2pir)`B. zeroC. `(mu_(0)n^(2)e)/(r)`D. `(mu_(0)"ne")/(2r)` |
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Answer» Correct Answer - d |
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| 513. |
An electron moving in a circular orbit of radius `r` makes `n` rotation per secound. The magnetic field produced at the centre has magnitudeA. zeroB. `(mu_0n^2e)/(r)`C. `(mu_0"ne")/(2r)`D. `(mu_0"ne")/(2pir)` |
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Answer» Correct Answer - C Current due to orbital motion of electron is `I=("charge")/("time period")=(e)/(1//n)=n e` Magnetic field due to circular motion of electron is `B=(mu_0)/(4pi)(2pinI)/(r)=(mu_0nI)/(2r)=(mu_0I)/(2r)` where n is number of loops. Here `n=1`, as electron is making one loop. `B=(mu_0)/(2r)("ne")=(mu_0"ne")/(2r)` |
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| 514. |
An electron moving in a circular orbit of radius `r` makes `n` rotation per secound. The magnetic field produced at the centre has magnitudeA. `(mu_(0)"ne")/(2r)`B. `(mu_(0)n^(2)e)/(2r)`C. `(mu_(0)"ne")/(2pir)`D. Zero |
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Answer» Correct Answer - A |
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| 515. |
A very long straight wire carries a current I. At the instant when a charge `+Q` at point `P` has velocity `vecV`, as shown, the force on the charge is A. Opposite to `OX`B. Along `OX`C. Opposite to `OY`D. Along `OY` |
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Answer» Correct Answer - D |
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| 516. |
A current carrying circular loop is freely suspended by a long thread. The plane of the loop will point in the directionA. Wherever left freeB. North-southC. East-westD. At `45^(@)` with the east-west direction |
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Answer» Correct Answer - C |
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| 517. |
A current carrying loop is free to turn in a uniform magnetic field.The loop will then come into equilibrium when its plane is inclined atA. `0^(@)` to the direction of the fieldB. `45^(@)`to the direction of the fieldC. `90^(@)` to the direction of the fieldD. `135^(@)` to the direction of the field |
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Answer» Correct Answer - C |
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| 518. |
A short bar magnet of moment `0*32JT^-1` is placed in a uniform external magnetic field of `0*15T`, if the bar is free to rotate in the plane of the field, which orientations would correspond to its, (i) stable and (ii) unstable equilibrium? What is the potential energy of the magnet in each case? |
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Answer» Here, `M=0*32JT^-1`, `B=0*15T` (i) In stable equilibrium, the bar magnet is aligned along the magnetic fiel, i.e. `theta=0^@` Potential Energy `=-MB cos 0^@=-0*32xx0*15xx0=-4*8xx10^-2J` (ii) In unstable equilibrium, the magnet is so oriented that magnetic moment is at `180^@` to the magnetic field i.e. `theta=180^@` Potential energy=`-MB cos 180^@=-0*32xx0*15(-1)=4*8xx10^-2J`. |
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| 519. |
A bar magnet of magnetic moment M is aligned parallel to the direction of a uniform magnetic field B. Calculate work done to align the magnetic moment (i) opposite to the field (ii) normal to field direction? |
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Answer» Work done in turning a dipole from orientation `theta_1` to `theta_2` is given by `W=-MB(cos theta_2-cos theta_1)` (i) In aligning `vecM` opposite to magnetic field, `theta_1=0`, `theta_2=180^@` `:. W=-MB ( cos 180^@-cos 0^@)` `=-MB(-1-1)=2MB` (ii) In aligning `vecM` perpendicular to magnetic field, `theta_1=0^@`, `theta_2=90^@` `W=-MB(cos 90^@-cos0^@)=+MB` |
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| 520. |
A bar magnet of magnetic moment `1.5JT^-1` lies aligned with the direction of a uniform magnetic field of `0.22T`. (a) What is the amount of work done to turn the magnet so as to align its magnetic moment (i) normal to the field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)? |
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Answer» Here, `M=1.5JT^-1`, `B=0.22T`, `W=?` (a) (i) Here, `theta_1=0^@` (along the field), `theta_2=90^@` (`_|_` to the field) As `W=-MB(costheta_2-costheta_1)` `:. W=-1.5xx0.22(cos 90^@-cos0^@)=-0.33(0-1)=0.33J` (ii) Here, `theta_1=0^@`, `theta_2=180^@` `W=-1.5xx0.22(cos 180^@-cos0^@)=-0.33(-1-1)=0.66J` (b) Torque `tau=MB sin theta` (i) Here, `theta=90^@`, `tau=1.5xx0.22sin90^@=0.33Nm` (ii) Here, `theta=180^@`, `tau=1.5xx0.22sin180^@=0` |
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| 521. |
A straight horizontal conducting rod of length `0*45m` and mass `60g` is suspended by two vertical wires at its end. A current of `5*0A` is set up in the rod through the wires. (a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero? (b) What will be the total tension in the wires if the direction of current is reversed, keeping the magnetic field same as before. (Ignore the mass of the wire) `g=9*8ms^-2`. |
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Answer» Here, `l=0*45m`, `m=60g=60xx10^-3kg`, `I=5*0A`. (a) Tension in the wire is zero if force on the wire carrying current due to magnetic field is equal and opposite to the weight of wire i.e. `BIl=mg` or `B=mg//Il=(60xx10^-3)xx9*8//(5*0xx0*45)=0*26T` The force due to magnetic field will be upwards if the direction of field is horizontal and normal to the conductor. (b) When direction of current is reversed, Bil and mg will act vertically downwards, the effective tension in the wires, `T=BIl+mg=0*26xx5*0xx0*45+(60xx10^-3)xx9*8=1*176N` |
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| 522. |
A magnetic field set up using Helmholtz coils (described in Question 16 above) is uniform in a small region and has a magnitude of `0*75T`. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through `15kV` enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is `9xx10^5Vm^-1`, make a simple guess so to what the beam contains. Why is the answer not unique? |
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Answer» Here, `B=0*75T`, `E=9xx10^5Vm^-1`, `V=15kV=15000V, e//m=?` Let e be the charge and m be the mass of the particles. Let v be the velocity acquired by the particles, when accelerated under an accelerating voltage V, then `1/2mv^2=eV` Since the particles are not deflected by the two crossed fields, so `eE=evB` or `v=E//B` `:. 1/2m(E^2)/(B^2)=eV` or `e/m=(E^2)/(2VB^2)=((9*0xx10^5)^2)/(2xx15000xx(0*75)^2)=4*8xx10^7C//kg` This value of `e//m(=4*8xx10^7C//kg)` corresponds to deutrons, hence particles are deutrons ions. But the above value of `e//m` is also for `He^(++)` and `Li^(+++)`. [As `e//m=2 e//2m=3 e//3m]`, so the particles can be `He^(+++)` or `Li^(+++)` also. |
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| 523. |
Calculate the magnetic field at the center of a regualr hexagon, 10 cm on each side, when the current through the wire forming the hexagon is 5A. |
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Answer» Correct Answer - 34.6 `muT` |
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| 524. |
Can we have a magnet with a single pole? |
| Answer» No, because unlike poles of equal strength exist together. | |
| 525. |
Are the two poles of a magnet equally strong? |
| Answer» Correct Answer - Yes, always. | |
| 526. |
Write an expression for magnetic field intensity at any point an axial line of a bar magnet. |
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Answer» `B_1=(mu_0)/(4pi)(2Md)/((d^2-l^2)^2)`, along SN where symbols have usual meaning. |
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| 527. |
In hydrogen atom, an electron is revolving in the orbit of radius `0.53 Å` with `6.6xx10^(15) rotations//second`. Magnetic field produced at the centre of the orbit isA. `0.125wb//m^(2)`B. `1.25wb//m^(2)`C. `12.5wb//m^(2)`D. `125wb//m^(2)` |
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Answer» Correct Answer - C |
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| 528. |
Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field `vec(B)=B_(0)hat(K)`A. They have equal z-components of momentaB. They must have equal chargesC. They necessarily represent a particle-antiparticle pairD. The charge to mass ratio satisfy: `(e/m)_1+(e/m)_2=0` |
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Answer» Correct Answer - D In a uniform magnetic field, the two charged particles will traverse identical helical paths in a completely opposite sense if the charge/mass ratio of these two particles is same and charges on them are of opposite character. In this situation `(e//m)_1+(e//m)_2=0`, holds good. |
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| 529. |
A short bar magnet has a magnetic moment of `0*48JT^-1`. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of `10cm` from the centre of the magnet on (i) the axis (ii) the equatorial line (normal bisector) of the magnet. |
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Answer» Here, `M=0.48JT^-1`, `B=?`, `d=10cm=0.1m` (i) On the axis of the magnet, `B=(mu_0)/(4pi)(2M)/(d^3)=10^-7xx(2xx0*48)/((0*1)^3)=0*96xx10^-4T` along S-N direction. (ii) On the equatorial line of the magnet `B=(mu_0)/(4pi)xx(M)/(d^3)=10^-7xx(0.48)/((0.1)^3)=0.48xx10^-4T`, along N-S direction. |
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| 530. |
Statement1: The magnetic filed at the ends of very long current carrying solenoid is half of that at the centre. Statement2: If the solenoid is sufficiently long, the field within it is uniform. |
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Answer» Correct Answer - `B` For a solenoid Bend `=(1)/(2) Bi n` also for a long solenoid magnetic field is uniform within it but this reason is not explaining thy statement (I) . |
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| 531. |
Why is soft iron preferred for making the core of a transformer? |
| Answer» The area of hysterisis loop for soft iron is very small. Therefore, energy dissipated per unit volume per cycle of magnetisation is small. | |
| 532. |
If the magnetic dipole of moment of an atom of diamagnetic material, paramagnetic material and ferromagnetic material are donated by `mu_d, mu_p` and `mu_f` respectively, then:A. `mu_p=0` and `mu_f!=0`B. `mu_d=!=0` and `mu_p=0`C. `mu_d!=0` and `mu_f!=0`D. `mu_d=0` and `mu_p!=0` |
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Answer» Correct Answer - D According to electron theory of magnetism, an atom of a diamagentic material has no intrinsic dipole moment, whereas atom of a para-magnetic material has some intrinsic dipole moment ltbr. i.e., `mu_d=0` and `mu_p!=0` |
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| 533. |
In a hydrogen atom, an electron moves in a circular orbit og radius `5.2xx10^(-11)m` and produces a magnetic induction of `12.56 T` at it necleus. The current produced by the motion of the electron will be (Given `mu_(0)=4pixx10^(-7) W//bA-m`)A. `6.53xx10^(-3)` ampereB. `13.25xx10^(-10)` ampereC. `9.6xx10^(6)` ampereD. `1.04xx10^(-3)` ampere |
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Answer» Correct Answer - D |
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| 534. |
Why are electromagnets made of soft iron? |
| Answer» This is because coercivity of soft iron is small. | |
| 535. |
Magnetic field strength, ………………., flux …………………represents the same quantity. |
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Answer» Correct Answer - magnetic induction ; density of magnetic field. |
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| 536. |
The relative magnetic permeability of a magnetic material is `800`. Identify the type of material. |
| Answer» A relative magnetic permeability is positive and high, it is ferromagnetic in nature. | |
| 537. |
Relative magnetic permeability of a material is defined as the ratio of ………………….to……………….of…………… . |
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Answer» Correct Answer - magnetic permeability of the material ; magnetic permeability ; free space. |
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| 538. |
The `(tau - theta)` graph for a current carrying coil placed in a uniform magnetic field isA. B. C. D. |
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Answer» Correct Answer - A |
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| 539. |
A wire carrying a current `i` is placed in a uniform magnetic field in the form of the curve `y=asin((pix)/L)0 lexle2L`. The force acting on the wire is A. `(iBL)/(pi)`B. `iBLpi`C. `2iBL`D. zero |
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Answer» Correct Answer - C |
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| 540. |
`3 A` of current is flowing in a linear conductor having a length of `40 cm`. The conductor is placed in a magnetic field of strength 500 gauss and makes an angle of `30^(@)` with the direction of the field. It experiences a force of magnitudeA. `3xx10^(4)` newtonB. `3xx10^(2)` newtonC. `3xx10^(-2)` newtonD. `3xx10^(-4)` newton |
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Answer» Correct Answer - C |
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| 541. |
Magnetic field intensity at the centre of coil of `50` turns, radius `0.5m` and carrying a current of `2A` isA. `0.5xx10^(-5)T`B. `1.25xx10^(-4)T`C. `3xx10^(-5)T`D. `4xx10^(-5)T` |
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Answer» Correct Answer - B |
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| 542. |
A current is flowing through a thin cylindrical shell of radius `R`. If energy density in the medium, due to magnetic field, at a distance `r` from axis of the shell is equal to `U` then which of the following graphs is correctA. B. C. D. |
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Answer» Correct Answer - B |
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| 543. |
If in circular coil of radius `R`, current `I` is flowing and in another coil `B` of radius `2R` a current `2I` is flowing , then the raatio of the magnetic fields `B_(A) and B_(B)`, produced by them will beA. `4:1`B. `2:1`C. `3:1`D. `1:1` |
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Answer» Correct Answer - D |
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| 544. |
If in circular coil of radius `R`, current `I` is flowing and in another coil `B` of radius `2R` a current `2I` is flowing , then the raatio of the magnetic fields `B_(A) and B_(B)`, produced by them will beA. `1`B. `2`C. `1//2`D. `4` |
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Answer» Correct Answer - 1 Magnetic field at centre of circular coil `A` is `B_(A) =(mu_(0)Ni)/(2R)` `R` is radius and I is current flowing in coil Similarly `B_(B) =mu_(0N(2i))/(2.(2R))` `=(mu_0Ni)/(2R) = (B_(A))/(B_(B)) =1` . |
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| 545. |
The magnetic field at a distance r from a long wire carryimg current I is 0.4 T. The magnetic field at a distance 2r isA. 0.2 TeslaB. 0.8 TeslaC. 0.1 TeslaD. 1.6 Tesla |
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Answer» Correct Answer - A |
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| 546. |
Figure shows a square loop ABCD with edge length `a`. The resistance of the wire ABC is `r` and that of ADC is `2r`. Find the magnetic field B at the centre of the loop assuming uniform wires. ` A. `(sqrt(2)mu_(0)i)/(3pia)o.`B. `(sqrt(2)mu_(0)i)/(3pia) ox`C. `(sqrt(2)mu_(0)i)/(pia) o.`D. `(sqrt(2)mu_(0)i)/(pi a) ox` |
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Answer» Correct Answer - B |
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| 547. |
If current flowing through shell of previous objective is equal to `i`, then energy density at a point distance `2R` from axis of the shell varies according to the graphA. B. C. D. |
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Answer» Correct Answer - B |
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| 548. |
Figure shows a square loop ABCD with edge length `a`. The resistance of the wire ABC is `r` and that of ADC is `2r`. Find the magnetic field B at the centre of the loop assuming uniform wires. ` A. `(2mu_0I)/(pir)`B. `(sqrt2mu_0I)/(pir)`C. `(sqrt2mu_0I)/(3pir)`D. `(2mu_0I)/(3pir)` |
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Answer» Correct Answer - C Let `I_1` and `I_2` be the currents in arms PQS and PTS respectively. Then `I=I_1+I_2`. As resistance of arm `PQS=R` and resistance of arm `PTS=2R` Therefore, `I_1=(2I)/(3)` and `I_2=I/3` Magnetic field induction at O due to current through PQ or QS is `B_(PQ)=B_(QS)=(mu_0)/(4pi)(I_1)/(r//2)[sin pi/4+sin pi/4]` `=(mu_0)/(4pi)(2I//3)/(r//2)xx2/sqrt2=(sqrtmu_0I)/(3pir)` acting perpendicular to the plane of loop downwards. Magnetic field induction at O due to current through PT or TS is `B_(PT)=B_(TS)=(mu_0)/(4pi)(I//3)/(r//2)(sin pi/4+sin pi/4)=(sqrt2mu_0I)/(6pir)` acting perpendicular to the plane of loop upwards. The resultant magnetic field induction at O is `B=B_(PQ)+B_(QS)-B_(PT)-B_(TS)` `=(sqrt2mu_0I)/(3pir)+(sqrt2mu_0I)/(3pir)-(sqrt2mu_0I)/(6pir)-(sqrt2mu_0I)/(6pir)` `=(sqrt2mu_0I)/(3pir)` |
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| 549. |
The magnetic field at a distance r from a long wire carryimg current I is 0.4 T. The magnetic field at a distance 2r isA. 0.2 TB. 0.8 TC. 0.1 TD. 1.6 T |
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Answer» Correct Answer - a |
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| 550. |
Figure shows a square loop made from a uniform wire. If a battery is connected between the pionts A and C. What will be the magnetic field at the centre O of the square? |
| Answer» At point A, the current I divides equally. So current in arm AB and BC is `I//2` and current in arm `AD` and `DC` is also `I//2`. The magnetic fields at O due to the current in wires AB and DC are equal and opposite. Also the magnetic fields at O due to currents in wires AD and BC are equal and opposite. Hence, the resultant magnetic field at O will be zero. | |