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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
11i don't know........
i don't have the solutions right now.
Well u can post the solution as well......
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
