Question
A square loop with $$2.0\,m$$ sides is perpendicular to a uniform magnetic field, with half the area of the loop in the field is shown in figure. The loop contains a $$20.0\,V$$ battery with negligible internal resistance. If the magnitude of the field varies with time according to $$B = 0.042 - 0.87\,t,$$ with $$B$$ in tesla and $$t$$ in second. The net emf of the circuit is:
A square loop with $$2.0\,m$$ sides is perpendicular to a uniform magnetic field, with half the area of the loop in the field is shown in figure. The loop contains a $$20.0\,V$$ battery with negligible internal resistance. If the magnitude of the field varies with time according to $$B = 0.042 - 0.87\,t,$$ with $$B$$ in tesla and $$t$$ in second. The net emf of the circuit is:
A.
$$20.0\,V$$
B.
$$18.26\,V$$
C.
$$21.74\,V$$
D.
None of these
Answer :
$$21.74\,V$$
Solution :
Induced emf. $$\left| e \right| = \frac{{d\phi }}{{dt}} = A\left( {\frac{{dB}}{{dt}}} \right)$$
$$\eqalign{ & = \frac{{2 \times 2}}{2} \times \frac{d}{{dt}}\left( {0.042 - 0.87\,t} \right) \cr & = 2 \times 0.87 = 1.74\,V \cr} $$
The net induced emf $$ = 20 + 1.74 = 21.74\,V$$
Induced emf. $$\left| e \right| = \frac{{d\phi }}{{dt}} = A\left( {\frac{{dB}}{{dt}}} \right)$$
$$\eqalign{ & = \frac{{2 \times 2}}{2} \times \frac{d}{{dt}}\left( {0.042 - 0.87\,t} \right) \cr & = 2 \times 0.87 = 1.74\,V \cr} $$
The net induced emf $$ = 20 + 1.74 = 21.74\,V$$
