Manipal Entrance Test 2015 Physics Paper

Current is given by I=

E

R′

Substituting, E=−

ndϕ

dt

, we get
I=−

n

R′

.

dϕ

dt

...(i)
Given that w1 and w2 are the values of the flux associated with one tum of the coil.
Therefore, Eq. (i) becomes,
I=−

n

R′

[

w2−w1

t2−t1

]....(ii)
Total resistance of the combination is R′=R+4R=5R, substituting the values of R′ and (t2−t1)=t in Eq (ii),we get
I=−

n

5R

(

w2−w1

t

)
⇒I=

−n(w2−w1)

5Rt