AP EAPCET 24-Aug-2021 Shift 2 Paper

Given that, number of turns, N=25
Radius of coil, r=1.8‌cm=1.8×10−2m
Resistance of coil, R=5Ω
Magnetic field,
B=(0.2t−0.05t2)T
Now, area of coil,
A=πr2=3.14×(1.8×10−2)2
=1.02×10−3m2
Induced emf, ε=

−Nd

dt

(BA)=−NA‌

dB

dt

=−NA‌

d

dt

(0.2t−0.05t2)
ε=−NA(0.2−0.1t)
Substituting the values, we get
ε=−25×1.02×10−3(0.2−0.1t)
=−2.55×10−2(0.2−0.1t)
Now, instantaneous power,
P=

ε2

R

=

[−2.55×10−2(0.2−0.1t)]2

5

At, t=35, power dissipated is
P=

(2.55×10−2)2(0.2−0.1×3)2

5

=1.30×10−6W=1.30µW
It is close to option (d).