The change in potential energy is -0.7065J
Given,
number of turns in the loop is N = 50
radius of the loop is r = 10cm = 0.1m
current through it i = 3A
the magnetic field is 0.3T
This question is solved by using the formula of magnetic potential energy.
Magnetic PE is defined as
U = -μ.B = -μBcosΘ
where μ is the magnetic moment and is defined as μ = NiA
where A is an area of the loop.
when the loop is at an angle of Θ with the magnetic field then its energy is
U₀ = -μBcos Θ
and when it is in a stable equilibrium position i.e it is parallel with the
magnetic field so that the angle Θ = 0° now U' = -μBcos Θ = -μB
so change in magnetic PE is
AU = U' - U₀ = -μB + μBcosΘ = μB(cos Θ - 1)
The magnetic PE is defined as U = - μ. B = -μBcos Θ
The magnetic moment of the loop is μ = NiA
substitute the values
A = 50 x (3A) X T X (0.1)2
A = 4.71Am²
so change in PE is
ΔU - μB(cos Θ - 1)
= - (4.71Am²) (0.3T)(cos 60 - 1)
= -0.7065J
negative sign shows that the dipole is in stable equilibrium(minimum PE).
The unit of magnetic potential energy is Joule (J).
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