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 \[I=5A\] Length of wire \[l=1\,cm=0.01\text{ }m\] Angle between the wires and magnetic field is \[\theta -90{}^\circ \] Magnetic field B = 0.8 T Now the force on each cm of wires is \[F=IlB\,\sin \theta \] \[=5\times 0.01\times 0.8\,\sin 90{}^\circ \] \[=0.04\,N\]You need to login to perform this action.
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