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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)

best_answer

 

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

Posted by

prateek

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