11i don't know........
i don't have the solutions right now.
Well u can post the solution as well......
Two coils of wire having identical resistances, R, are connected as shown
the upper end of the first coil is connected to the end of the second and vice-versa. their cross-sectional areas are A1 and A2. the two are kept in a uniform magnetic field of induction B. the pair is suddenly rotated about their common axis by 180°. The charge tranferred to the circuit in the process equals:-
a) B(A1 + A2) / R
b) 2B(A1 + A2)/ R
c) B(A1 - A2)/R
D) 2B(A1 - A2)/R
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8 Answers
11
1i given yesterday in narayana full test-1
process if u can understand
Q=l Δφl/R
flux linked to final position BA1-BA2
and original configuration is BA2-BA1
Δφ=2B(A1-A2)
thinking u have understood for better understanding make direction of i in both loop & write flux linked with each loop taking one direction as positive
11well rahul i too gave the narayana full test but didn't attempted this one and i also didn't collected the solution book afterwards.
thats why???
1dont know but i got the sol. but of no use for me bcoz many ans. r given wrong
