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…

Question

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 axis of rotation of the coil. The maximum voltage generated in the coil is (2π)/(β) volt. The value of β is ______.

MARKS Content Team · Accepted Answer

Correct answer: 5. Time taken for one complete revolution = time period =2 s. Therefore, ω=(2π)/T=π rad s^(-1) Now flux, ϕ=NABcos(ωt) Therefore, induced EMF ε=-(dϕ)/dt=NABωsin(ωt) The maximum voltage generated, ε_max=NABω =200×0.2×0.01×π =(4π)/10=(2π)/5 volt Hence, β=5.

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 axis of rotation of the coil. The maximum voltage generated in the coil is $\frac{2 π}{β}$ volt. The value of $β$ is ______.

Solution

Relevant Questions

A coil of 200 turns and area 0.20 m2 is rotated at half a revolution per second and is placed in uniform magnetic field of 0.01 T perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is 2πβ volt. The value of β is ______.

Solution

Relevant Questions

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 axis of rotation of the coil. The maximum voltage generated in the coil is $\frac{2 π}{β}$ volt. The value of $β$ is ______.