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प्रश्न
Figure 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.
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उत्तर
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,
\[i = \frac{B_v lv}{R}\]
\[\Rightarrow B_v = \frac{iR}{lv}\]
`=(2xx10^-5xx2xx10^-1)/(20xx10^-2xx20xx10^-2)=(2xx2xx10^-7)/(2xx2xx10^-2)`
\[= 1 \times {10}^{- 5} T\]
We know,
\[\tan\delta = \frac{B_v}{B_H} = \frac{1 \times {10}^{- 5}}{3 \times {10}^{- 5}} = \frac{1}{3}\]
\[ \Rightarrow \delta = \tan^{- 1} \left( \frac{1}{3} \right)\]
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