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A loop of flexible conducting wire of le...

A loop of flexible conducting wire of length 0.5 m lies in a magnetic field of 1.0 T perpendicular to the plane of the loop. Show that when a current as shown in Fig. 1.112 is passed through the loop, it opens into a circle. Also calculate the tension developed in the wire if the current is 1.57 A.

Text Solution

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At every element of length `dl` of the current carrying loop placed in a magnetic field, a force `IB` `dl` will act, whose direction is perpendicular to the element. Hence the loop will take the circular from figure.

Let an arc element CD(=length `dl`), of the loop subtend an angle `phi` at the centre O of the circle (of radius r) and T be the tension at each of its ends. Then components of tension taken along AO, must be due to force acting on the current element due to magnetic field. So
`2T sin phi//2=IBdl` or `2Tphi/2=IB dl`
[ `:'` For small angle, `sin phi/2=phi/2`]
or `T=(IBdl)/(phi)=IBr` (`:' phi=(dl)/(r)`)
`=IB(l)/(2pi)=(1*57xx1*0xx0*5)/(2xx3*14)=0*125N`
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