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<div>A copper wire of density d with cross section area S bent to make three sides of as square can turn about a horizontal axis OO' (FIGURE). The wire is located in uniform vertcal magntic field. Find the magnetic field induction if on passing a current I through the wire the latter deflects by an angle&nbsp;&theta; .</div> <div>&nbsp;</div> <div>&nbsp;</div> <div>&nbsp;</div> <div>&nbsp;</div> <div>&nbsp;</div> <div>&nbsp;</div> <div><img src="https://images.topperlearning.com/topper/tinymce/imagemanager/files/cd47d3d2ad34085f9a2b751a0b36f3625559f6b3b7a566.67210764Qu.jpg" alt="" /></div>
Asked by Sourav | 18 May, 2015, 07:57: PM

A copper wire of density d with cross section area S bent to make three sides of as square can turn about a horizontal axis OO'

The equal forces established on the sides OP and OP’ are directed along the same line in opposite direction. Therefore, torque acting on the side and its net force about the axis OO’ is zero.  The force on PP’ is deflecting in nature; hence, restoring torque developed due the weight of the shape must be equal to deflecting torque.

Let l be the length of each side.

From the figure, in equilibrium

ilb(l cosθ) = (Sld)g ½ sin θ + (Sld)g ½ sin θ + (Sld)g sin θ

il² B cosθ = 2Sdgl² sinθ

Answered by Priyanka Kumbhar | 19 May, 2015, 11:28: AM

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