A straight wire 15cm long, carrying a current of 6.0A is in a uniform field of 0.40T. What is the force on the wire when it is at right angle to the field
- 0.46N
- 0.35N
-
0.36N
- 0,24N
The correct answer is: C
Explanation
F = BIL = 0.4 x 6 x \(\frac{15}{100}\)
= 0.36N
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