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The self-inductance of a coil having 400 turns is 10 mH. The magnetic flux through the cross-section of the coil corresponding to current 2 mA is(a) 4 x 10−3 Wb(b) 3 x 10−3 Wb(c)2 x 10−3 Wb(d) 8 x 10−3 Wb |
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Answer» The self-inductance of a coil having 400 turns is 10 mH. The magnetic flux through the cross-section of the coil corresponding to current 2 mA is (a) 4 x 10−3 Wb (b) 3 x 10−3 Wb (c)2 x 10−3 Wb (d) 8 x 10−3 Wb |
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