A circular loop is prepared from a wire of uniform cross section. A battery is connected between any two points on its circumference. The magnetic induction at the centre of the loop is ______ .

Explanation

A battery is joined between points A and B of the loop as shown in the figure.
Since the cross-section of the wire is uniform, the resistance of the part of wire is proportional to the length of that part. `(becauseR=rhol/A)`
Let the resistance per unit length be R..
Length of wire ACB = `l_(1)`
Length of wire ADB = `l_(2)`
`therefore` Resistance of wire ACB = `R_(1)=R.l_(1)`
Resistance of wire ADB = `R_(2)=R.l_(2)`
Current in wire ACB = `I_(1)`
Current in wire ADB = `I_(2)`
These two parts ACB and ADB are in parallel between A and B points.
`V=I_(1)R_(1)=I_(2)R_(2)`
`I_(1)(R.l_(1))=I_(2)(R.l_(2))`
`thereforeI_(1)l_(1)=I_(2)l_(2)" "...(A)`
Every small current element of this wire is perpendicular to the position vector of O, with respect to it.
`therefore` Biot-Savart.s law gives, magnetic field at O due to ACB, as
`B_(1)=mu_(0)/(4pi)(I_(1)l_(1)sin90^(@))/r^(2)`
and that due to ADB,
`B_(2)=mu_(0)/(4pi)(I_(2)l_(2)sin90^(@))/r^(2)`
Since, `I_(1)l_(1)=I_(2)l_(2)`
we get, `B_(1)=B_(2)`.
According to right hand rule, the directions of `B_(1)andB_(2)` are opposite to each other. Hence the resultant magnetic field at O will be zero.