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Find the magnetic force on a short magnet of magnetic dipole moment M_(2) due to another short magnet of magnetic dipole moment M_(1). |
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Answer» Solution :To find the magnetic force we will use the formula of `B` due to a magnet.We will also assume `m` and `-m` as pole strengths of `N` and `S` of `M_(2)`.Also length of `M_(2)` as `2a.B_(1)` and `B_(2)` are the strengths of the magnetic field due to `M_(1)` at `+m` and `-m` respectively.They experience magnetic forces `F_(1)` and `F_(2)` as shown. `F_(1)=2(mu_(0)/(4pi))M_(1)/(r-a)^(3)m` and `F_(2)=2(mu_(0)/(4pi))M_(1)/(r-a)^(3)m` `:. F_(res=F_(1)-F_(2)=2((mu_(0))/(4pi))M_(1)m[(1/(r-a)^(3))-(1/(r+a)^(3))]` `=2((mu_(0))/(4pi))(M_(1)m)/r^(3)[(1-a/r)^(-3)-(1+a/r)^(-3)]` By using acceleration, Binomial expansion, and neglecting TERMS of high power we get `F_(res)=2((mu_(0))/(4pi))(M_(1)m)/r^(3)[1+(3a)/r-1+(3a)/r]` `=2((mu_(0))/(4pi))(M_(1)m)/r^(3)(6a)/r` `=2((mu_(0))/(4pi))(M_(1)3M_(2))/r^(4)` `=6((mu_(0))/(4pi))(M_(1)M_(2))/r^(4)` Direction of `F_(res)` is towards RIGHT. Alternative Method `=B((mu_(0))/(4pi))(2M_(1))/r^(3)rArr (dB)/(dr)=-(mu_(0))/(4pi)XX(6M_(1))/r^(4)` `F=-M_(2)xx(dB)/(dr) rArr F=((mu_(0))/(4pi))(6M_(1)M_(2))/r^(4)` 
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