1Vab = VA - Vb
hmm..
emf of OA - emf of OB ??
a conducting wire is bent in d form of a loop as shown.
the segments AOB is parabolic given by the eqn y^2 =2x while
segment BA is a straight line parallel to the y-axis.
the magnetic field in the region is B= - 8 k^
and the current in the wire is 2A.
now answer the follwoing qs.
1) the torque on the loop is:
a) 16√2 Nm b) 16Nm c)18√2Nm d)zero.
2) the field created by the current in the loop at point C will be:
a) -μ0/4Πk^ b) -μ0/2Πk^
c) -μ0√2 /Πk^ d) none of these.
3.) if the loop were rotated abt the z-axis with an angular velocity of
1 rad/sec then magnitude of emf induced across the straight segment AC will be:
a) 32V b) 8V c) 16V d) 4V
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46 Answers
33Yes :)
Isko integrate kar ke bhi dekh liye...(VAC wala..)wahi aata hai.. :)
33Zero hi hai...
waise ek baar bana lena (integrations se...) to acha rahega...:
13i didnt get how is the ans for 1st one d....can sum1 plz help...
33One more good extension to it...
Find potential difference between A and B....
33B is there as earlier...and same ω
Then it is zero...
This is for any rod..
A
|
0---------|
|
B
rotating about O ... then ant symmetric points (A and B) its zero...
33P Q ka coordinates bahut acha di ho... :P
P(2,-1),, Q bhi (2,-1) [3]
13ya saw it...logic of imagining a rod of length 2 samaj nahi aya...
i cud do it integrate karke.....
1if u din like dat method....
do like...
V(OA) - V(OC) ...
Y INTEGRATE.. ??
its lenthy r8?
