A) \[100\text{ }A\]
B) \[10\text{ }A\]
C) \[1\text{ }A\]
D) zero
Correct Answer: B
Solution :
Force on a current carrying wire of length l, carrying current i, kept in a magnetic field B is given by \[\vec{F}=i\,\vec{l}\times \vec{B}=iBl\,\sin \theta \] where \[\theta \] is the angle the wire makes with the magnetic field. Given, \[l=1m,\,\theta ={{90}^{o}},\,\,F=1kg-wt=9.8N,\] \[B=0.98T\] \[\therefore \] \[i=\frac{{\vec{F}}}{lB}=\frac{9.8}{0.98\times 1}=10A\]
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