Moving Charges And Magnetism

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Question
CBSEENPH12040010

A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and the other in x-z plane.  if the current in the loop is i. The resultant magnetic field due to the two semicircular parts at their common centre is

  • fraction numerator straight mu subscript straight o space straight i over denominator 2 square root of 2 straight R end root end fraction
  • fraction numerator straight mu subscript 0 straight i over denominator 2 straight R end fraction
  • fraction numerator straight mu subscript straight o straight i over denominator 4 straight R end fraction
  • fraction numerator straight mu subscript straight o straight i over denominator square root of 2 straight R end fraction

Solution

A.

fraction numerator straight mu subscript straight o space straight i over denominator 2 square root of 2 straight R end root end fraction WiredFaculty

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Question
CBSEENPH12039984

A galvanometer has a coil of resistance 100 Ω and gives a full-scale deflection for 30 mA current. If it is to work as a voltmeter of 30 V range, the resistance required to be added will be

  • 900 Ω

  • 1800

  • 500 Ω

  • 1000 Ω

Solution

A.

900 Ω

Required resistance
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Question
CBSEENPH12040065

A galvanometer of resistance 50 Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

  • 5050 Ω

  • 5550 Ω

  • 6050 Ω

  • 4450 Ω

Solution

D.

4450 Ω

Current through the galvanometer 

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Question
CBSEENPH12039961

A galvanometer of resistance, G is shunted by a resistance S ohm. To keep the main current in the circuit unchanged,the resistance to be put in series with the galvanometer is

  • fraction numerator straight S squared over denominator left parenthesis straight S space plus straight G right parenthesis end fraction
  • fraction numerator SG over denominator left parenthesis straight S plus straight G right parenthesis end fraction
  • fraction numerator straight G squared over denominator left parenthesis straight S space plus straight G right parenthesis end fraction
  • fraction numerator straight G over denominator left parenthesis straight S plus straight G right parenthesis end fraction

Solution

C.

fraction numerator straight G squared over denominator left parenthesis straight S space plus straight G right parenthesis end fraction

Current will be unchanged if resistance remains same so

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Question
CBSEENPH12040058

A long solenoid has 500 turns. When a current of 2 A is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 x 10-3 Wb. The self -inductance of the solenoid is 

  • 2.5 H

  • 2.0 H

  • 1.0 H

  • 4.0 H

Solution

C.

1.0 H

The inductance of a coil is numerically equal to the emf induced in the coil when the current in the coil changes at the rate of 1 As-1. If I is the current flowing in the circuit, then flux linked with the circuit is observed to be proportional to I, ie,
WiredFaculty
where L is called the self-inductance or coefficients of self-inductance or coefficient of self-inductance or simply inductance of the coil.
Net flux through the solenoid.
Φ = 500 x 4 x 10-3 = 2 Wb
Or 2 = L x 2 [after putting value in eq. or  L = 1H]

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