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A metallic rod of length $$20$$ cm is placed in North-South direction and is moved at a constant speed of $$20$$ m s$$^{-1}$$ towards East. The horizontal component of the Earth's magnetic field at that place is $$4 \times 10^{-3}$$ T and the angle of dip is $$45°$$. The emf induced in the rod is ______ mV.
Correct Answer: 16
Given:
l=20 cm=0.2 m, v=20 m/s
Step 1: Vertical component of Earth’s field
$$B_V=B_H\tanθ=B_H(\sin ce\ θ=45∘)$$
Step 2: Induced emf
$$\varepsilon=(\vec{v}\times\vec{B})\cdot\vec{l}\ =B_Vvl$$
Step 3: Calculation
V$$\varepsilon=(4\times10^{-3})(20)(0.2)$$
= $$16\times10^{-3}V$$
= $$=16mV$$
Final Answer:
16 mV
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