A current carrying wire AC is placed in uniform transverse magnetic field then the force on wire AC:-

3 N

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4.2 N

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6 N

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4 N

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Solution

The correct option is B 3 N
$S o l u t i o n :$ $A C = 10 c m = \frac{1}{10} m$
$\begin{matrix} F & = I l B sin θ θ = 90^{\circ} = 3 \times \frac{1}{10} \times 10 = 3 N \end{matrix}$
$S o, t h e$ $c o r r e c t$ $o p t i o n : A$

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A current carrying wire AC is placed in uniform transverse magnetic field then the force on wire AC:-

The correct option is B 3 NSolution: AC=10cm=110mF=IlBsinθθ=90∘=3×110×10=3NSo,the correct option:A

The correct option is B 3 N
$S o l u t i o n :$ $A C = 10 c m = \frac{1}{10} m$
$\begin{matrix} F & = I l B sin θ θ = 90^{\circ} = 3 \times \frac{1}{10} \times 10 = 3 N \end{matrix}$
$S o, t h e$ $c o r r e c t$ $o p t i o n : A$