JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 9)
A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity $$\omega ,$$ the maxium e.m.f. induced in the coil will be:
3 nBA$$\omega $$
$${3 \over 2}$$ nBA$$\omega $$
nBA$$\omega $$
$${1 \over 2}$$ nBA$$\omega $$
설명
Flux in the coil, $$\phi $$ = nBA sin($$\omega $$t)
When n = no. of turns
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ A = Area of coil
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$$$\omega $$ = angular speed
Induced emf,
$$\left| e \right| = {{d\phi } \over {dt}}$$
= nBA$$\omega $$ cos$$\omega $$t
$$\therefore\,\,\,$$ emax = nBA$$\omega $$
When n = no. of turns
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ A = Area of coil
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$$$\omega $$ = angular speed
Induced emf,
$$\left| e \right| = {{d\phi } \over {dt}}$$
= nBA$$\omega $$ cos$$\omega $$t
$$\therefore\,\,\,$$ emax = nBA$$\omega $$