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Question

A wire loop PQRS formed by joining two semi-circular wires of radii R1 and R2 carries a current I as shown in the following diagram. The magnetic induction at the centre O is

311187_cef88fcd985c481f83b3e84dc7596855.png

A
μ0I4R1
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B
μ0I4R2
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C
μ04πI(1R11R2)
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D
μ04I(1R1+1R2)
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Solution

The correct option is A μ04πI(1R11R2)
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 = μ0I2πR, where R is the radius of the loop.

The magnetic field at the center of half a current loop B = 12μ0I2R=μ0I4R.

Therefore,

B1 = μ0I4R1 and B2 = - μ0I4R2.

So, B = B1+B2 = μ0I4R1 - μ0I4R2.

That is, μ0I4(1R11R2).

Hence, the magnetic induction at the center is μ0I4(1R11R2).

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