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,

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

$$\Rightarrow B_v=\frac{iR}{lv}$$

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

$$=1\times {10}^{-5}T$$

We know,

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

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