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Two conducting wires are joined together at an angle ‘\mathrm {2\theta}’ and placed in the vertical magnetic field of intensity Bo (see figure. A slide wire of mass ‘M’ and having resistance per unit length as ‘k’ is moved along the conducting wires with constant
velocity ‘vo’ with a variable force ‘P’. Find the variation of ‘P’ with time, so that the slide wire can move with constant velocity vo. (Initially the slide wire was placed at origin and had velocity vo)

Option: 1

\frac{\mathrm{B}_0^2 v_0^2 x \tan \theta(t)}{k}


Option: 2

\frac{\mathrm{B}_{\mathrm{o}}^2 \mathrm{v}_{\mathrm{O}}^2 \mathrm{x} \tan \theta(\mathrm{t})}{2 \mathrm{k}}


Option: 3

\frac{2 \mathrm{~B}_{\mathrm{o}}^2 \mathrm{v}_{\mathrm{O}}^2 \mathrm{x} \tan \theta(\mathrm{t})}{\mathrm{k}}


Option: 4

\frac{\mathrm{B}_{\mathrm{o}}^2 \mathrm{v}_{\mathrm{O}}^2 \mathrm{x} \tan \theta(\mathrm{t})}{3 \mathrm{k}}


Answers (1)

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Let slide wire has moved through a distance ‘x’ then MN = 2x
\tan \theta at the resistance of the circuit =\mathrm{k}(2 \mathrm{x} \tan \theta) .........(1)
\phi_{\mathrm{B}}=\mathrm{B}_{\mathrm{o}} \mathrm{x}^2 \tan \theta\\

and  \mathrm{t} \frac{-\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}=-2 \mathrm{~B}_0 \mathrm{v}_{\mathrm{o}} \mathrm{x} \tan \theta

Current in the circuit is \mathrm{'i'}=\frac{2 \mathrm{~B}_{\mathrm{o}} \mathrm{v}_{\mathrm{o}} \mathrm{x} \tan \theta}{\mathrm{k}(2 \mathrm{x} \tan \theta)}=\frac{\mathrm{B}_{\mathrm{o}} \mathrm{v}_{\mathrm{o}}}{\mathrm{k}}.................(2)

.So then slide wire moves with uniform velocity, the ‘P’ needed to motion is given by

\mathrm{B}_{\mathrm{o}} (i) (2 x \tan \theta)\\ Or, \quad \mathrm{P}=\mathrm{B}_{\mathrm{o}}\left(\frac{\mathrm{B}_{\mathrm{o}} \mathrm{V}_{\mathrm{o}}}{\mathrm{k}}\right) 2 \mathrm{x} \tan \theta=\frac{2 \mathrm{~B}_{\mathrm{o}}^2 \mathrm{v}_{\mathrm{o}}}{\mathrm{k}} \mathrm{x} \tan \theta\\ \mathrm {Also, x=v_0 t} \\\mathrm {So, \quad P=\frac{2 B_0 v_0}{k(2 x \tan \theta)}=\frac{B_0^2 v_0^2 x \tan \theta(t)}{k}}

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Devendra Khairwa

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