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A straight wire of length 2m carries a current of 10A. If this wire is placed in a uniform magnetic field of 0.15T making an angle of 45^(@) with the magnetic field, the applied force on the wire will be |
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Answer» 1.5N `|VEC(F)| = |I (vec(l) xx vec(B))| or F= IlB SIN theta` Here, I=10A, l=2M, B= 0.15T, `theta= 45^(@)` `:.` Force, `F= (10A) (2m) (0.15T) sin 45^(@) = (3)/(sqrt2)N` |
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