Question

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 strength 3.96 × 10-3 newtons/amp·meter ?

Answer 1

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

Answer 2

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