Question
A magnetic field of $$2 \times {10^{ - 2}}T$$ acts at right angles to a coil of area $$100\,c{m^2},$$ with 50 turns. The average emf induced in the coil is $$0.1\,V,$$ when it is removed from the field in $$t$$ second. The value of $$t$$ is
A.
$$10\,s$$
B.
$$0.1\,s$$
C.
$$0.01\,s$$
D.
$$1\,s$$
Answer :
$$0.1\,s$$
Solution :
Emf induced in the coil due to change in magnetic flux
$$e = - \frac{{d\phi }}{{dt}} = - \frac{{\left( {{\phi _2} - {\phi _1}} \right)}}{{dt}}$$
When magnetic field is perpendicular to coil
$${\phi _1} = NBA$$
When coil is removed, $${\phi _2} = 0$$
So, $$e = - \frac{{\left( {0 - NBA} \right)}}{{dt}}\,\,{\text{or}}\,\,dt = \frac{{NBA}}{e}$$
Here, $$N = 50,B = 2 \times {10^{ - 2}}T,A = 100\,c{m^2}$$
$$\eqalign{
& = {10^{ - 2}}{m^2}\,{\text{and}}\,\,e = 0.1\,V \cr
& \therefore dt = \frac{{50 \times 2 \times {{10}^{ - 2}} \times {{10}^{ - 2}}}}{{0.1}} \cr
& = 0.1\,s \cr} $$