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?

Question

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?

Astro · Answer

Welcome to QuestionCove.

The velocity= 4.20 M/S

EMF= .88V

SOMF= B=.075 N/A-m

So to find the EMF we must....EMF=L*B*V

.88/0.075*4.20=2.79m

If we round to the nearest tenth from the number above we get a total length of the wire in the magnetic field which is 2.8m

(M=Meters)
(EMF=Electrical Magnetic Field)
(SOMF=Strength of Magnetic Feild)

Astro · Answer

@astro wrote:

Welcome to QuestionCove. The velocity= 4.20 M/S EMF= .88V SOMF= B=.075 N/A-m So to find the EMF we must....EMF=L*B*V .88/0.075*4.20=2.79m If we round to the nearest tenth from the number above we get a total length of the wire in the magnetic field which is 2.8m (M=Meters) (EMF=Electrical Magnetic Field) (SOMF=Strength of Magnetic Feild)

Sorry...... L = EMF/B*v = 0.88/0.075*4.20 = 2.79 m

Astro · Answer

Also forgot to explain the it is magnetic field B, so...

The equation would be.... Length=Electromagnetic field/ Strength of Magnetic field, B*Velocity=.88/0.05*4.20=2.79

Astro · Answer

Tried to explain it the best I could~