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An Intiative by Sri Chaitanya
a
IaB2
b
IaB2
c
IaB32
d
I a B3
answer is A.
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Detailed Solution
Force on TP=-Floop =IaBsin90°+60°=IaB2⊙ Floop =IaB2⊗AlternativelyFPQ=IaBsin90°+30°⊗=IaB32⊗ FQP=I(2a)Bsin30°⊗=IaB⊗ FRS=I(a)Bsin60°⊗=IaB32⊙ FST=IaBsin90°+60⊙=IaB2⊙ So, Fnet =IaB2⊗