In an electric motor, wires carrying a current of $5 A$ are placed at right angles to a magnetic field of induction $0.8 T$ . If each wire has length of $20 c m$ , then the force acting on each wire is :

0.2 N

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0.4 N

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0.6 N

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0.8 N

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Solution

The correct option is D $0.8 N$
$F = B i l = 0.8 \times 5 \times 20 \times 10^{- 2} = 0.8 N$

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In an electric motor, wires carrying a current of 5A are placed at right angles to a magnetic field of induction 0.8T. If each wire has length of 20cm, then the force acting on each wire is :

The correct option is D 0.8NF=Bil=0.8×5×20×10−2=0.8N

In an electric motor, wires carrying a current of $5 A$ are placed at right angles to a magnetic field of induction $0.8 T$ . If each wire has length of $20 c m$ , then the force acting on each wire is :

The correct option is D $0.8 N$
$F = B i l = 0.8 \times 5 \times 20 \times 10^{- 2} = 0.8 N$