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Question

Two wire loops ABCD formed by joining two semicircular wires of radii $${R}_{1}$$ and $${R}_{2}$$ carries a current I as shown in figure.The magentic induction at the centre is
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A
μ0I(R2R1)/4R1R2 out of the page
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B
μ0I(R2R1)/4R1R2 into the page
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C
μ0IR1R24(R2R1) out of the page
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D
Zero
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Solution

The correct option is A $${\mu}_{0}I({R}_{2}-{R}_{1})/4{R}_{1}{R}_{2}$$ out of the page
Answer is A.
The resultant magnetic field at O will be the sum of the magnetic fields due to the current in the two semicircles, and we can use the expression for the magnetic field at the center of a current loop to find B.
The resultant magnetic field at point O is given as B = B1+B2
The magnetic field at the center of a current loop is B = $$\dfrac { { \mu  }_{ 0 }I }{ 2\pi R } $$, where R is the radius of the loop.
The magnetic field at the center of half a current loop B = $$\dfrac { 1 }{ 2 } \dfrac { { \mu  }_{ 0 }I }{ 2R } =\dfrac { { \mu  }_{ 0 }I }{ 4R } $$.
Therefore,
B1 = $$\dfrac { { \mu  }_{ 0 }I }{ 4R1 } $$ and B2 = - $$\dfrac { { \mu  }_{ 0 }I }{ 4R2 } $$.
So, B = B1+B2 = $$\dfrac { { \mu  }_{ 0 }I }{ 4R1 } $$ - $$\dfrac { { \mu  }_{ 0 }I }{ 4R2 } $$.
That is, $$\dfrac { { \mu  }_{ 0 }I }{ 4 } (\dfrac { 1 }{ R1 } -\dfrac { 1 }{ R2 } )\quad =\quad { \mu  }_{ 0 }I(\dfrac { R2-R1 }{ 4R1.R2 } )$$.
Hence, the magnetic induction at the center is $${ \mu  }_{ 0 }I(\dfrac { R2-R1 }{ 4R1.R2 } )$$ out of the page.

Physics

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