A coil with 25 turns of wire moves in a uniform magnetic fieldof 1.5 T. The plan

Question

A coil with 25 turns of wire moves in a uniform magnetic fieldof 1.5 T. The plane of the magnetic field is perpendicular to theplane of the coil. The coil has a cross-sectional area of 0.80m2. The coil exits the field in 1.0 s. If thecoil’s resistance is 2.0 W, what is the inducedcurrent? A coil with 25 turns of wire moves in a uniform magnetic fieldof 1.5 T. The plane of the magnetic field is perpendicular to theplane of the coil. The coil has a cross-sectional area of 0.80m2. The coil exits the field in 1.0 s. If thecoil’s resistance is 2.0 W, what is the inducedcurrent?

Explanation / Answer

   InducedEMF      E   =   -d / dt   =   - N *(B2   -   B1) *A / t    N   =   No. ofturns in thecoil   =   25,   B1   =   Initialmagneticfield   =   1.5   T    B2   =   Finalmagenticfield   =   0,   A   =   Areaof crosssection   =   0.80   m2,   t   =   time   =   1.0   s    =>   E   =   -25 * ( 0 - 1.5) * 0.80 /1.0   =   30.0   V    Inducedcurrent   I   =   V /R   =   30.0 /2.0   =      15.0   A

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