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 Multiple Choice QuestionsMultiple Choice Questions

1.

A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orientations as shown in the figures below:



If there is a uniform magnetic field of 0.3 T in the positive z direction , in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium?

  • (a) and (b) respectively

  • (a) and (c) respectively

  • (b) and (d) respectively

  • (b) and (c)  respectively


C.

(b) and (d) respectively

Since, B is uniform only torque acts on a current carrying loop

As, straight tau space equals space bold M bold space bold x bold space bold B
vertical line straight tau vertical line space equals space vertical line bold M vertical line vertical line bold B vertical line space sin space straight theta
For space orientation space shown space in space left parenthesis straight b right parenthesis space straight theta space equals space 0 to the power of 0 comma space straight tau space equals space 0 space stable space equilibrium

443 Views

2.

Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper as shown. The variation of the magnetic field B along the line XX' is given by


B.

The magnetic field in between because of each of the conductors will be in naturally opposite directions.
∴ Net magnetic field

straight B subscript in space between space space end subscript space equals space fraction numerator straight mu subscript straight o straight i over denominator 2 space straight pi space straight x end fraction space bold j with bold hat on top space plus space fraction numerator straight mu subscript straight o straight i over denominator 2 space straight pi space left parenthesis 2 straight d minus straight x right parenthesis end fraction left parenthesis negative straight j right parenthesis
space equals space fraction numerator straight mu subscript straight o straight i over denominator 2 space straight pi space straight x end fraction space open square brackets 1 over straight x minus fraction numerator 1 over denominator 2 straight d minus straight x end fraction close square brackets space left parenthesis space bold j with bold hat on top space right parenthesis
At x = d, B in between = 0
For x < d, B in between =bold j with bold hat on top
For x> d, B in between = left parenthesis bold minus bold j with bold hat on top right parenthesis
Towards x, net magnetic field will add up and direction will be  left parenthesis bold minus bold j with bold hat on top right parenthesis. Towards' x' b net magnetic field will add up and direction will be (bold j with bold hat on top)

544 Views

3.

An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to:

  • 80 H

  • 0.08 H

  • 0.044 H

  • 0.065 H

D.

0.065 H

I = 10 A, V = 80 V
R = V/I = 80/10 = 8Ω and ω = 50 Hz

For AC circuit, we have 




straight I space equals space fraction numerator straight V over denominator square root of 8 squared plus straight X subscript straight L superscript 2 end root end fraction
rightwards double arrow space 10 space equals space fraction numerator 220 over denominator square root of 64 plus straight X subscript straight L superscript 2 end root end fraction
rightwards double arrow space square root of 64 space plus straight X subscript straight L superscript 2 end root space equals space 22
squaring space on space both space sides comma space we space get
64 space plus space straight X subscript straight L superscript 2 space equals space 484
straight X subscript straight L superscript 2 space equals space 484 minus 64 space equals space 420
straight X subscript straight L space equals space square root of 420
2 straight pi space straight x space ωL space equals space square root of 420
Series space induction space on space an space arc space lamp.
straight L space equals space fraction numerator square root of 420 over denominator left parenthesis 2 straight pi space straight x space 50 right parenthesis end fraction space equals space 0.065 space straight H

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4.

A galvanometer having a coil resistance of 100 Ω gives a full-scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full-scale deflection for a current of 10 A, is:

  • 0.01 Ω

  • 2 Ω

  • 0.1 Ω

  • 3 Ω


A.

0.01 Ω

Maximum voltage that can be applied across the galvanometer coil = 100 Ω x 10-3 A = 0.1



If Rs is the shunt resistance, then
Rs x 10 A = 0.1 V
Rs = 0.01 Ω

192 Views

5.

A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB andCD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30º. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the origin.


Due to the presence of the current I1 at the origin


The magnitude of the magnetic field (2) due to the loop ABCD at the origin (O) is

  • The forces on AB and DC are zero

  • The forces on AD and BC are zero

  • The magnitude of the net force on the loop is given by fraction numerator straight mu subscript straight o II subscript 1 over denominator 4 straight pi end fraction space open square brackets 2 left parenthesis straight b minus straight a right parenthesis space plus space straight pi over 3 left parenthesis straight a plus straight b right parenthesis close square brackets

  • The magnitude of the net force on the loop is given by fraction numerator straight mu subscript 0 space II subscript 1 over denominator 24 space ab end fraction space left parenthesis straight b minus straight a right parenthesis


B.

The forces on AD and BC are zero

The forces on AD and BC are zero because magnetic field due to a straight wire on AD and BC is parallel to elementary length of the loop.

191 Views

6.

A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB andCD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30º. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the origin.



The magnitude of the magnetic field (2) due to the loop ABCD at the origin (O) is

  • zero

  • fraction numerator straight mu space left parenthesis straight b minus straight a right parenthesis over denominator 24 ab end fraction
  • fraction numerator straight mu subscript straight o straight I over denominator 4 space straight pi end fraction space open square brackets fraction numerator straight b minus straight a over denominator ab end fraction close square brackets
  • fraction numerator straight mu subscript 0 straight I over denominator 4 straight pi end fraction space open square brackets 2 left parenthesis straight b minus straight a right parenthesis space plus straight pi over 3 left parenthesis straight a plus straight b right parenthesis close square brackets

B.

fraction numerator straight mu space left parenthesis straight b minus straight a right parenthesis over denominator 24 ab end fraction
straight B space equals space 1 over 12 space of space fraction numerator straight mu subscript 0 straight I over denominator 2 end fraction space open parentheses 1 over straight a minus 1 over straight b close parentheses
space equals space fraction numerator straight mu subscript 0 straight I space left parenthesis straight b minus straight a right parenthesis over denominator 24 space ab end fraction
125 Views

7.

Hysteresis loops for two magnetic materials A and B are given below:



These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use:

  • A for electric generators and transformers.

  • A for electromagnets and B for electric generators

  • A for transformers and B for electric generators.

  • B for electromagnets and transformers.


D.

B for electromagnets and transformers.

Area of the hysteresis loop is proportional to the net energy absorbed per unit volume by the material, as it is taken over a complete cycle of magnetisation.

For electromagnets and transformers, energy loss should be low.

i.e thin hysteresis curves.

Also |B|→0 When H = 0 and |H| should be small when B →0.

387 Views

8.

In a coil of resistance 100 Ω, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is

  • 250 Wb

  • 275 Wb

  • 200 Wb

  • 225 Wb


A.

250 Wb



straight q space equals space fraction numerator increment straight ϕ over denominator straight R end fraction
increment straight ϕ space equals space change space in space flux
straight q space equals space integral Idt
equals space Area space of space current space minus time space graph
space equals 1 half space straight x space 10 space straight x space 0.5 space equals space 2.5 space coloumb
straight q space equals fraction numerator increment straight ϕ over denominator straight R end fraction
increment straight ϕ space equals space 2.5 space straight x space 100 space equals space 250 space space wb
944 Views

9.

Two coaxial solenoids of different radii carry current I in the same direction. Let stack straight F subscript 1 with rightwards arrow on top be the magnetic force on the inner solenoid due to the outer one and stack straight F subscript 2 with rightwards arrow on top be the magnetic force on the outer solenoid due to the inner one. Then:

  • stack straight F subscript 1 space with rightwards arrow on top equals stack straight F subscript 2 space with rightwards arrow on top equals space 0
  • stack straight F subscript 1 with rightwards arrow on top is radially inwards and stack straight F subscript 2 with rightwards arrow on top is radially outwards
  • stack straight F subscript 1 with rightwards arrow on top is radially inwards and stack straight F subscript 2 with rightwards arrow on top =0
  • stack straight F subscript 1 with rightwards arrow on top is radially outward and stack straight F subscript 2 with rightwards arrow on top =0

A.

stack straight F subscript 1 space with rightwards arrow on top equals stack straight F subscript 2 space with rightwards arrow on top equals space 0

Consider the two coaxial solenoids. Due to one of the solenoids magnetic field at the centre of the other can be assumed to be constant


Due to symmetry, forces on upper and lower part of the solenoid will be equal and opposite and hence resultant is zero.
Therefore option (a) is correct.

1119 Views

10.

Two identical wires A and B, each of length ‘l’, carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side ‘a’. If BA and BB are the values of the magnetic field at the centres of the circle and square respectively, then the ratio BA /BB is:

  • straight pi squared over 8
  • fraction numerator straight pi squared over denominator 16 square root of 2 end fraction
  • fraction numerator straight pi squared over denominator 8 square root of 2 end fraction

C.

Magnetic field in case of circle of radius R, we have



straight B subscript straight A space equals space fraction numerator straight mu subscript straight o straight I over denominator 2 straight R end fraction
As comma space 2 πR space equals space straight l space left parenthesis straight l space is space length space of space straight a space wire right parenthesis
straight R space equals fraction numerator straight l over denominator 2 straight pi end fraction
straight B subscript straight A space equals space fraction numerator straight mu subscript straight o straight I over denominator 2 space straight x space begin display style fraction numerator straight l over denominator 2 straight pi end fraction end style end fraction space equals space fraction numerator straight mu subscript straight o Iπ over denominator straight l end fraction space... space left parenthesis straight i right parenthesis

Magnetic field in case of square of side we get



281 Views