When a wheel with metallic spokes 1 2m long rotates in a magnetic field of 5 * 10-5 T perpendicular - Physics - Electromagnetic Induction

Question

When a wheel with metallic spokes 1 2m long rotates in a
magnetic field of 5 * 10-5 T perpendicular to the plane
of the wheel , an emf of 10-2V is induced across the rim
and the axle of the wheel Find the rate of rotation of the
wheel

Vijay Yadav · Accepted Answer

The emf between the ends of rotating spoke is $ξ = \int d ξ = \int_{0}^{r} B ω x d x = \frac{1}{2} B ω r^{2}$ Given that B = 5 Ã— 10â€“ 5 T $ξ = 10^{- 2} V r = 1 2 m ω = ?$ $ω = \frac{2 ξ}{B r^{2}} = \frac{2 \times 10^{- 2}}{5 \times 10^{- 5} \times (1 2)^{2}} = \frac{2 \times 10^{3}}{5 \times 1 44} = \frac{2000}{5 \times 1 44} = 277 78 rad/s ω = 2 π ν = 2 77 78 rad/s ν = \frac{277 78}{2 π} = 44 20 rot/s$

When a wheel with metallic spokes 1.2m long rotates in a magnetic field of 5 * 10-5 T perpendicular to the plane of the wheel , an emf of 10-2V is induced across the rim and the axle of the wheel . Find the .rate of rotation of the wheel

When a wheel with metallic spokes 1.2m long rotates in a magnetic field of 5 * 10^-5 T perpendicular to the plane of the wheel , an emf of 10^-2V is induced across the rim and the axle of the wheel . Find the .rate of rotation of the wheel