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Moving Charges And Magnetism

Question
CBSEENPH12040087
Wired Faculty App

Two circular coils 1 and 2 are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil. What is the ratio of potential difference applied across them so that the magnetic field at their centres is the same?

  • 3

  • 4

  • 6

  • 2

Solution

B.

4

Magnetic field at the centre of a circular coil is  
                           straight B space equals space fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction cross times fraction numerator 2 πi over denominator straight r end fraction
where i is current flowing in the coil and r is radius of coil.
At the centre of coil - 1,
                straight B subscript 1 space equals space fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction cross times fraction numerator 2 πi subscript 1 over denominator straight r subscript 1 end fraction space space space space space space space space space... left parenthesis straight i right parenthesis
At the centre of coil-2
                           straight B subscript 2 equals space space fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction cross times fraction numerator 2 πi subscript 2 over denominator straight r subscript 2 end fraction space space space space space space space... left parenthesis ii right parenthesis
But                   straight B subscript 1 space equals space straight B subscript 2
therefore space space space space fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction space fraction numerator 2 πi subscript 1 over denominator straight r subscript 1 end fraction space equals space fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction fraction numerator 2 πi subscript 2 over denominator straight r subscript 2 end fraction or space space space space space space space space straight i subscript 1 over straight r subscript 1 space equals space straight i subscript 2 over straight r subscript 2 As space space space space space space space space straight r subscript 1 space equals space 2 straight r subscript 2 therefore space space space space space space fraction numerator straight i subscript 1 over denominator 2 straight r subscript 2 end fraction space equals space straight i subscript 2 over straight r subscript 2 or space space space space space space space space straight i subscript 1 space equals space 2 straight i subscript 2 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis iii right parenthesis space space space space space space
Now, export of potential differences
                straight V subscript 2 over straight V subscript 1 space equals space fraction numerator straight i subscript 2 cross times straight r subscript 2 over denominator straight i subscript 1 cross times straight r subscript 1 end fraction space equals space fraction numerator straight i subscript 2 cross times straight r subscript 2 over denominator 2 straight i subscript 2 cross times 2 straight r subscript 2 end fraction space equals space 1 fourth
therefore space space space space space straight V subscript 1 over straight V subscript 2 space equals space 4 over 1

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M1 and M2 have the following particulars:
R1 = 10 Ω   N1 = 30,
A1 = 3.6 x 10–3 m2,  B1 = 0.25 T
R2 = 14 Ω;  N2 = 42,
A2 = 1.8 x 10–3 m2, B2 = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

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