SolveItJEE Main
Physics · 2024

JEE Main · 1 February 2024, Shift 2 · Q58

A coil of 200 turns and area 0.20 m^2 is rotated at half a revolution per second and is placed in uniform magnetic field of 0.01 T perpendicular to…

A coil of $\displaystyle 200$ turns and area $\displaystyle 0.20 \mathrm{~m}^2$ is rotated at half a revolution per second and is placed in uniform magnetic field of $\displaystyle 0.01$ T perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is $\displaystyle \frac{2 \pi}{\beta}$ volt. The value of $\displaystyle \beta$ is $\displaystyle \_\_\_\_$.
ShareWhatsAppTelegram

More from Electromagnetic Induction

JEE Main 2024 Physics question, with the answer from NTA’s final answer key. Where our answers come from.