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,

$i=\frac{B_{v}lv}{R}$

$\Rightarrow B_{\mathrm{v}}=\frac{iR}{lv}$

$=\frac{2\times {10}^{-5}\times 2\times {10}^{-1}}{20\times {10}^{-2}\times 20\times {10}^{-2}}=\frac{2\times 2\times {10}^{-7}}{2\times 2\times {10}^{-2}}$

$=1\times {10}^{-5}\mathrm{T}$
We know,
$$\begin{gathered} \tan \delta =\frac{B_{\mathrm{v}}}{B_{\mathrm{H}}}=\frac{1\times {10}^{-5}}{3\times {10}^{-5}}=\frac{1}{3} \\ \Rightarrow \delta ={\tan }^{-1}\left(\frac{1}{3}\right) \end{gathered}$$