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

A
μ0I(R2R1)/4R1R2 out of the page
B
μ0I(R2R1)/4R1R2 into the page
C
μ0IR1R24(R2R1) out of the page
D
Zero

Solution

## The correct option is A $${\mu}_{0}I({R}_{2}-{R}_{1})/4{R}_{1}{R}_{2}$$ out of the pageAnswer 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+B2The 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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