A 20 m long horizontal wire extends from North-East to South-West. It is falling down with a speed of 4 m/s horizontally. The horizontal components of the earth's magnetic field is \( 0.4\times10^{-4}\,\mathrm{T} \). Find the value of induced current in the wire if the resistance of wire is 2Ω is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.13 mA
A wire falling with speed v through a magnetic field has a motional emf equal to B v times the length of wire perpendicular to B (v is perpendicular to B_H here). The vertical component of the earth's field is parallel to the velocity and gives no emf, so only the horizontal component \( B_H \) acts, and \( B_H \) points north.
The wire lies along North-East to South-West, so it makes \( 45^\circ \) with the north direction. The length of wire perpendicular to \( B_H \) is \( l\sin45^\circ \).
\( \varepsilon = B_Hv\,l\sin45^\circ = 0.4\times10^{-4}\,\mathrm{T}\times4\,\mathrm{m\,s^{-1}}\times20\,\mathrm{m}\times\frac{1}{\sqrt{2}} \)
\( \varepsilon = 3.2\times10^{-3}\times0.707\,\mathrm{V} \approx 2.26\times10^{-3}\,\mathrm{V} \)
\( \displaystyle I = \frac{\varepsilon}{R} = \frac{2.26\times10^{-3}\,\mathrm{V}}{2\,\Omega} \approx 1.13\times10^{-3}\,\mathrm{A} = 1.13\,\mathrm{mA} \)