1.

A conductor of length 50 cm carrying a current of 5 A is placed perpendicular to a magnetic field of induction 2 × 10-3 T. Find the force on the conductor

Answer»

Force on the conductor = ILB 

= 5 × 50 × 10-2 × 2 × 10-3 

= – 5 × 10-3 N



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