Question 28

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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