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VITEEE 2009 Paper

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Question : 31 of 120

Marks: +1, -0

A wire of length

l

is bent into a circular loop of radius R and carries a current

I

. The magnetic field at the centre of the loop is B. The same wire is now bent into a double loop of equal radii. If both loops carry the same current

I

and it is in the same direction, the magnetic field at the centre of the double loop will be

Zero
2B
4B
8B

Solution:

At the centre of the loop, magnetic field

B=\frac{\mu_0}{4\pi}\cdot\frac{I\cdot 2\pi R}{R^2}

...(i)

For the wire which is looped double let radius becomes r

Then,

\frac{l}{2}=2\pi r

or

\frac{l}{4\pi}=r

∴

B'=\frac{\mu_0}{4\pi}\cdot\frac{I\cdot 2\pi r\times 2}{r^2}

or

B'=\frac{\mu_0}{4\pi}\cdot\frac{I\cdot\frac{l}{2}\cdot 2}{\left(\frac{1}{4\pi}\right)^2}

or

B'=\frac{\mu_0}{4\pi}\cdot\frac{I l\times 16\pi^2}{l^2}

...(ii)

Now,

B=\frac{\mu_0}{4\pi}\cdot\frac{I\cdot l}{\left(\frac{l}{2\pi}\right)^2}\left[R=\frac{1}{2\pi}\right]

...(iii)

Dividing Eq. (ii) by (iii),

\frac{B'}{B}=\frac{\frac{\mu_0}{4\pi}\cdot\frac{I\cdot l\cdot 16\pi^2}{l^2}}{\frac{\mu_0}{4\pi}\cdot\frac{I l\cdot 4\pi^2}{l^2}}=4

or B' = 4B

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