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