Question

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.

Answer 1

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, B_H = 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 B_v 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,

\[\tan\delta = \frac{B_v}{B_H} = \frac{1 \times {10}^{- 5}}{3 \times {10}^{- 5}} = \frac{1}{3}\]

\[ \Rightarrow \delta =  \tan^{- 1} \left( \frac{1}{3} \right)\]