InterviewSolution
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InterviewSolution
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An atom in the state `.^(2)P_(3//2)` is located on the axis of a loop of radius `r=5 cm` carrying a current, `I=10 A`. The distance between the aotm and the centre of the loop is equal to the radius of the latter. How great may be the maximum force that the magnetic field of that current exerts on the atom? |
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Answer» The force on an atom with magnetic moment `vec(mu)` in a magnetic field of induction `vec(B)` is given by `vec(F)=(vec(mu).vec(grad))vec(B)` In the present case, the maximum force arise when `vec(mu)` is along the axis of close to it. Then`F_(Ƶ)=(mu_(Ƶ))_(max)(del B)/(delƵ)` Here `(mu_(Ƶ))_(max)=gmu_(B)J`. The Lande factor `g` is for `^(2)P_(1//2)` `g=1+((1)/(2)xx(3)/(2)+(1)/(2)xx(3)/(2)-1xx2)/(2(1)/(2)xx(3)/(2))=1-(1//2)/(3//2)=(2)/(3)` and `J=(1)/(2) so (mu_(Z))_(max)=(1)/(3)mu_(B)`. The magnetic field is given by `B_(Ƶ)=(mu_(0))/(4 pi).(2Ipi r^(2))/((r^(2)+Ƶ^(2))^(3//2))` or `(del B_(Ƶ))/(delƵ)=-(mu_(0))/(4pi)6I pir^(2)(Ƶ)/((r^(2)+Ƶ^(2))^(5//2))` Thus `((del B_(Ƶ))/(delƵ=r))=(mu_(0))/(4 pi)(3I pi)/(sqrt(8)r^(2))` Thus maximum force is `F=(1)/(3)mu_(B)(mu_(0))/(4 pi)(3 pi)/(sqrt(8))(I)/(r^(2))` Substitution gives (using data in `MKS` units) `F=4.1xx10^(-27)N` |
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