A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be

  • Option 1)

    nB

  • Option 2)

    n2B

  • Option 3)

    2nB

  • Option 4)

    2n2B

 

Answers (1)

 

Magnetic field due to cCrcular Current Carrying arc -

B=frac{mu_{o}}{4pi}:frac{2pi i}{r}=frac{mu_{o}i}{2r}

- wherein

 

 Let  L be the length of the wire

At the centre of loop B=\frac{\mu_{o} I}{2R}

B=\frac{\mu_{o} \pi I}{L}   \therefore L=2 \pi R

B^{1}=\frac{\mu_{o} I}{2r} = \frac{\mu_{o} I}{2\frac{L}{2n\pi}} \Rightarrow B^{1}=\frac{\mu_{o} n^{2}\pi I}{L}

\therefore Ratio \frac{B^{1}}{B}=n^{2}\Rightarrow B^{1}=n^{2}B


Option 1)

nB

This is incorrect option

Option 2)

n2B

This is correct option

Option 3)

2nB

This is incorrect option

Option 4)

2n2B

This is incorrect option

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