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Using the relation vec(B) = mu(0) (vec(H...

Using the relation `vec(B) = mu_(0) (vec(H)+ vec(M))`, show that `X_(m) = mu_(r) - 1`.

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`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) rArr X_(m) + 1 = (mu)/(mu_(0)) = mu_(r) `
`rArr X_(m) = mu_(r) - 1`
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