Question

A wire cuts across a flux of 2×(10)-2 weber in 0.12 seconds . what is the e.m.f induced in the wire​

Answer 1

Answer:

0.167 V

Explanation:

EMF=-d∅/dt

|EMF| = d∅/dt

∴EMF= 2 × 10⁻² / 0.12

= 2×10⁻²/(12/100)

=2×10⁻²×10² / 12

=1/6

=0.167 V

Answer 2

Given that,

dФ=$2*10^{-2}$

dt=0.12 seconds

∈=$-\frac{dФ}{dt}$

⇒|∈|=$\frac{2*10^{-2} }{0.12}$

=0.166 V

Hence the e.m.f induced in the wire is 0.166V