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If the emf produced in a wire is 0.88 volts and the wire moves perpendicular to a magnetic field of strength 0.075 newtons/amperes·meter at a speed of 4.20 meters/second, what is the length of the wire
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Nov 17, 2018
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If the emf produced in a wire is 0.88 volts and the wire moves perpendicular to a magnetic field of strength 0.075 newtons/amperes·meter at a speed of 4.20 meters/second, what is the length of the wire in the magnetic field?
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Anton Rodzik
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emf = -d(flux)/dt flux = B A, field times area normal to field B is constant but A is changing, as L sweeps along, dA/dt = L (4.20 m/s) emf = - B dA/dt
Vitomd
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Nov 19, 2018
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emf = d (phi-B) / dt
B dA/dt, where dA/dt is the area swept out by the wire per unit time.
0.88 V = (0.075 N/(A m)) (L)(4.20 m/s), so
L = (0.88 J/C) / [ (0.075 N s/C m)(4.2 m/s) ] = about 3 meters
IWheelBuy
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Nov 23, 2018
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