A coil carrying a current `10mA` is placed in a uniform magnetic field so that its axis coincides with the field direction. The single layer winding of the coil is made of copper wire with diameter `0*1mm`, radius of turns is equal to `30mm`. At what value of the induction of the external magnetic field can the coil winding be ruptured? Breaking stress is `3*1xx10^8Nm^-2`.

Explanation

Take a small element of length `dl` of the coil carrying current I. It is placed perpendicular to the magnetic field induction `vecB` acting upwards. It experience a force `Bidl`. Due to this force there will be tension T in the wire. If `dalpha` is the angle formed by the element at the centre O, then for equilibrium,

`BIdl=2(T sin dalpha//2)=2Tdalpha//2=Tdalpha`
[`:' d alpha` is small, hence `sin d alpha//2=d alpha//2`]
`:. T=(BIdl)/(dalpha)=BIr` `[:' r=dl//dalpha]`
Now stress on wire, `S=(T)/((piD^2//4))=(BIr)/((piD^2//4))`
If `B_m` is the maximum value of magnetic fied induction then breaking stress for rupture of wire is
`S_m=(B_mIr)/((piD^2)//4)` or `B_m=(piD^2S_m)/(4Ir)`
`:. B_m=22/7xx((0*1xx10^-3)^2xx(3*1xx10^8))/(4xx(10xx10^-3)xx(30xx10^-3))`
`=8*1+10^3T`