Answer : The emf produced is, 0.0368 volt
Solution :
Formula used :
[tex]emf=\frac{d\phi}{dt}=B\frac{dA}{dt}=B\times l\times v[/tex]
This equation is valid when B, l and v are mutually perpendicular to each other.
where,
B = magnetic field of strength = [tex]3.96\times 10^{-3}N/amp.meter[/tex]
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.
[tex]emf=B\times l\times v[/tex]
[tex]emf=(3.96\times 10^{-3}N/amp.meter)\times (1.5m)\times (6.2m/s)=0.0368volt[/tex]
Therefore, the emf produced is, 0.0368 volt
