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valentinak56 [21]
3 years ago
7

what is the emf produced in a 1.5 meter wire moving at a speed of 6.2 meters/second perpendicular to a magnetic field of strengt

h 3.96 × 10-3 newtons/amp·meter ?
Physics
2 answers:
ruslelena [56]3 years ago
7 0

Answer : The emf produced is, 0.0368 volt

Solution :

Formula used :

emf=\frac{d\phi}{dt}=B\frac{dA}{dt}=B\times l\times v

This equation is valid when B, l and v are mutually perpendicular to each other.

where,

B = magnetic field of strength = 3.96\times 10^{-3}N/amp.meter

l = length of wire = 1.5 m

v = velocity = 6.2 m/s

Now put all the given values in the above formula, we get the emf.

emf=B\times l\times v

emf=(3.96\times 10^{-3}N/amp.meter)\times (1.5m)\times (6.2m/s)=0.0368volt

Therefore, the emf produced is, 0.0368 volt

Svetlanka [38]3 years ago
5 0
General formula for emf is: emf=vBL(sin θ) ...(1) As the angle here is 90° and sin90°=1. So,equation (1) becomes; emf=vBL Putting values; emf=(6.2)(1.5)(3.96 × 10^-3)=0.0369 volts
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