Question

A ring of resistance $10\Omega$ , radius $10cm$ and $100$ turns is rotated $100$ revolutions per second about a fixed axis which is perpendicular to a uniform magnetic field of induction $10mT$ . The amplitude of the current in the loop will be nearly (Take ${\pi }^2=10$)

• A. $200A$
• B. $2A$
• C. $0.002A$
• D. None of these

Answer 1

The correct option is D None of these

The induced current is given as

$I=\frac{NBA\omega }{R}\sin \omega t$

The amplitude of the current is given as

$I=\frac{NBA\omega }{R}$

By substituting the given values in the above equation we get

$I=\frac{100\times 1\times {10}^{-3}T\times 3.14\times {\left(10\times {10}^{-2}\right)}^2\times 2\times 3.14\times 100}{10\Omega }$

$I=\frac{100\times 1\times {10}^{-3}T\times 10\times {\left(10\times {10}^{-2}\right)}^2\times 2\times 100}{10\Omega }$

$I=0.1A$