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The horizontal component of earth's magnetic field at a place is 3.5 10-5 T. A very long straight conductor carrying current of 2 A in the direction from South east to North West is placed. The force per unit length experienced by the conductor is 10-6 N/m.
35
[figure]
BH points north; current SE → NW makes =45 with it.
Fl=IBH = 2 3.5 10-5 1 2
=3.5 10-5=35 10-6 N/m
Answer: 35
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