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A straight conductor 2m long carrying of 15 A is kept at right angles to a uniform magnetic field of induction 5xx10^(-3) (Wb)//m^2. What is the force acting upon it ? |
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Answer» SOLUTION :We have `F=ilBsintheta` Here i=15amp, l=2m, `B=5xx10^(-3) (WB)//m^2` and `theta=90^@` `THEREFORE F=15xx2xx5xx10^(-3)=0.15N` |
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