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Question

Three long, straight and parallel wires C, D and G carrying currents are arranged as shown in Fig. The force experienced by a 25 cm length of wire C is


A

0.4 N

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B

0.04 N

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C

4×103N

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D

4×104N

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Solution

The correct option is D

4×104N


The magnetic field due to wire D at wire C is
BD=(μ04π)2Ir=107×2×300.03=2×104T
which is directed into the page.
Similarly, the field due to wire G at C is
BG=107×2×200.1=0.4×104 T
which is directed out of the page.
Therefore, the field at the position of the wire C is
B=BDBG=2×1040.4×104=1.6×104T
and is directed into the page.
The force on 25 cm of wire C is
F = BIl 90=1.6×104×10×0.25
=4×104N


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