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प्रश्न
A conductor of length 50 cm carrying a current of 5 A is placed perpendicular to a magnetic field of induction 2×10 -3T. Find the force on the conductor.
टीपा लिहा
उत्तर
Given: conductor of length = 50 cm
Current I= 5 A
Magnetic induction B = 2×10 -3T
Solution:
Force on the conductor = ILB
= 5 × 50 × 10-2 × 2 × 10-3
= 5 × 10-3
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