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Figure shows a square loop ABCD with edg...

Figure shows a square loop ABCD with edge length `a`. The resistance of the wire ABC is `r` and that of ADC is `2r`. Find the magnetic field B at the centre of the loop assuming uniform wires. `

A

`sqrt(2)mu_(0)i/3pi a` `o.`

B

`sqrt(2)mu_(0)i/3pi a` `ox`

C

`sqrt(2)mu_(0)i/pi a` `o.`

D

`sqrt(2)mu_(0)i/pi a` `ox`

Text Solution

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The correct Answer is:
2
According to question resistance of wire ADC is twice that of wire ABC. Hence current flows through ADC is half that of ABC i.e.`i_(2)/i_(1)=1/2.`Also i_(1)+i_(2)=I rArri_(1)=2i/3 and i_(2)=i/3` Magnetic field at centre O due to wire AB and BC (part 1 and 2) `B_(1) =mu_(0)/4pi .2i_(1)sin 45^@/a//2 ox` = mu_(0)/4pi .2sqrt(2)i_(1)/a ox`
and magnetic field at centre O due to wires AD and DC (i.e. part 3 and 4) `B_(3) = B_(4) = mu_(0)/4pi 2sqrt(2)i_(2)/a o.`
Hence net magnetic field at centre O, `B_(net) =(B_(1) +B_(2))-(B_(3) +B_(4))`
`=mu_(0)/4pi .4sqrt(2)i/3a (2-1) ox = sqrt(2)mu_(0)i/3pi a`
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