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  • A metallic rod of length is placed in North-South direction and

A metallic rod of length \mathrm{20 \mathrm{~cm}} is placed in North-South direction and is moved at a constant speed of \mathrm{20 \mathrm{~m} / \mathrm{s}} towards East. The horizontal component of the Earth's magnetic field at that place is \mathrm{4 \times 10^{-3} \mathrm{~T}} and the angle of dip is \mathrm{45^{\circ}}. The emf  induced in the rod is ________\mathrm{\mathrm{mV}}.

Option: 1

16


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer



\mathrm{B_{V}= B_{H}}
Vertical component \mathrm{\left ( B_{V} \right )} will produced motional emf.
\mathrm{E= B_{V} \times V\times l}
      \mathrm{= 4\times 10^{-3}\times 20\times 20\times10^{-2} }
      \mathrm{= 16\times 10^{-3}= 16\, mV }

The answer is 16

Posted by

Nehul

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