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Using the relation vec(B) = mu_(0) (vec(H)+ vec(M)), show that X_(m) = mu_(r) - 1. |
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Answer» Solution :`vec(B) = mu_(0) (vec(H) + vec(M))` But from EQUATION `x_(m) = |(vec(M))/(vec(H))|`, in vector form `vec(M) = X_(m) vec(H)` HENCE, `vec(B) = mu_(0) (X_(m) + 1)vec(H) rArr vec(B) = mu vec(H)` Where, `mu = mu_(0) (X_(m) + 1) rArrX_(m) + 1 = (mu)/(mu_(0)) = mu_(r) ` `rArr X_(m) = mu_(r) - 1` |
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