A) 0.08 N
B) 0.06 N
C) 0.04 N
D) 0.02 N
Correct Answer: C
Solution :
Here: current in the wire \[I=5\] amp Length of the wire = 1cm = 0.01 cm Angle between the wire and magnetic field\[\text{= 9}{{\text{0}}^{\text{o}}}\] Magnetic field B = 0.8 tesla Now, force on each cm of wire \[=IlB\sin \theta \] \[=5\times 0.01\times 0.8\sin {{90}^{o}}\] \[=0.04\times 1=0.04\,N\]
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