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The horizontal component of the earth's magnetic field at a certain place is 4.0xx10^(-5)T and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is east to west |
Answer» Solution :(a)  When I amount of current is passing through a straight conducting wire of length `vecl` (WHOSE direction is taken as the direction of current) placed in a uniform MAGNETIC field `VECB`, magnetic force `vecF` EXERTED on it is given by Ampere.s law as, `vecF=I(veclxxvecB)` `thereforeF=IlBsintheta` (where `theta` = angle between `veclandvecB`) Magnetic force per unit length, `F/l=IBsintheta""...(1)` = `IB_(h)sin90^(@)" "(because"Here "B=B_(h)andtheta=90^(@))` = `(1)(3xx10^(-5))(1)` = `3xx10^(-5)Nm^(-1)` (b) Now, in this case as per the statement, conducting wire is placed along `vecB_(h)` and current is passed from south to north and so in equation (1), we should take `theta=0^(@)`. Hence, in this case, `F/l=0" "(becausetheta=0^(@)rArrsintheta=0)` |
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