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

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

Answer 2

Answer : The emf produced is, 0.0368 volt

Solution :

Formula used :

`emf=(d\phi)/(dt)=B(dA)/(dt)=B* l* v`

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

where,

B = magnetic field of strength =
`3.96* 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* l* v`

`emf=(3.96* 10^(-3)N/amp.meter)* (1.5m)* (6.2m/s)=0.0368volt`

Therefore, the emf produced is, 0.0368 volt