NEET-XII-Physics
38: Electromagnetic Induction
- #40Figure shows a metallic wire of resistance 0.20 Ω sliding on a horizontal, U-shaped metallic rail. The separation between the parallel arms is 20 cm. An electric current of 2.0 µA passes through the wire when it is slid at a rate of 20 cm s-1. If the horizontal component of the earth’s magnetic field is 3.0 × 10-5 T, calculate the dip at the place.
FigureAns : Given:
Separation between the parallel arms, l = 20 cm = 20 × 10-2 m
Velocity of the sliding wire, v = 20 cm/s = 20 × 10-2 m/s
Horizontal component of the earth's magnetic field, BH = 3 × 10-5 T
Current through the wire, i = 2 µA = 2 × 10-6 A
Resistance of the wire, R = 0.2 Ω
Let the vertical component of the earth's magnetic field be Bv and the angle of the dip be δ.
Now,
`` i=\frac{{B}_{v}lv}{R}``
`` ``
`` \Rightarrow {B}_{\,\mathrm{\,v\,}}=\frac{iR}{lv}``
`` =\frac{2\times {10}^{-5}\times 2\times {10}^{-1}}{20\times {10}^{-2}\times 20\times {10}^{-2}}=\frac{2\times 2\times {10}^{-7}}{2\times 2\times {10}^{-2}}``
`` =1\times {10}^{-5}\,\mathrm{\,T\,}``
We know,
`` \,\mathrm{\,tan\,}\delta =\frac{{B}_{\,\mathrm{\,v\,}}}{{B}_{\,\mathrm{\,H\,}}}=\frac{1\times {10}^{-5}}{3\times {10}^{-5}}=\frac{1}{3}``
`` \Rightarrow \delta ={\,\mathrm{\,tan\,}}^{-1}\left(\frac{1}{3}\right)``
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