Question #330234

A closely wound 0.20m × .20 m square coil of 100 turns is exposed to an external magnetic field of 0.74 T that is perpendicular to the plane of the coil. The coil is then rotated for 10 s until it makes 60 degrees with the magnetic field. Find the average induced electromotive force during the 10 s interval.


1
Expert's answer
2022-04-18T17:31:21-0400

The emf induced in the coil is given by the Faraday's law:


E=ΔΦΔt\mathcal{E} = -\dfrac{\Delta\Phi}{\Delta t}

where ΔΦ\Delta\Phi is the change in magnetic flux, and Δt=10s\Delta t = 10s is the time elapsed. The initial magnetic flux through the coil is:


Φi=NBA\Phi_i = NBA

where N=100N = 100, B=0.74T,A=0.20m×0.20m=0.04m2B= 0.74T, A = 0.20m × 0.20 m=0.04m^2. The final flux is:


Φf=NBAcosθ\Phi_f = NBA\cos\theta

where θ=60°\theta=60\degree. Thus, the change is:


ΔΦ=ΦfΦi=NBA(cosθ1)\Delta\Phi = \Phi_f - \Phi_i = NBA(\cos\theta -1)

and the emf:


E=NBA(1cosθ)ΔtE=1000.740.04(1cos60°)10=0.148V\mathcal{E} = \dfrac{NBA(1-\cos\theta)}{\Delta t}\\ \mathcal{E} = \dfrac{100\cdot 0.74\cdot 0.04\cdot (1-\cos60\degree)}{10} = 0.148V

Answer. 0.148V.


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