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Derive the expression for magnetic force on a conductor carrying current keptin a magnetic field. |
Answer» Solution : Consider a conducting rod of LENGTH l and uniform area of cross-section A in an EXTERNAL magnetic field `vecB`as SHOWN in fig. When a steady current .I.flows in the rod, the TOTAL charge of mobile charge carriers is Q=n A l a The magnetic force on these charge carriers is `vecF = Q (vecV_d xx vecB) "" V_d to ` drift velocity `vecF = nAl q(vecV_d xx vecB) `.....(1) But `I = n q A V_d`......(2) Equation (2) in (1) we get `vecF = I(VECL xx vecB)` |
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