A magnetic dipole with a dipole moment of magnitude 0.0243 J/T is released from rest in a uniform magnetic field of magnitude 57
2 answers:
Answer:
The initial angle between the dipole moment and the magnetic field is 47.76⁰
Explanation:
Given;
magnitude of dipole moment, μ = 0.0243 J/T
magnitude of magnetic field, B = 57.5 mT
change in kinetic energy, ΔKE = 0.458 mJ
ΔKE = - ΔU
ΔKE = - (U₂ -U₁)
ΔKE = U₁ - U₂
U₁ -U₂ = 0.458 mJ
where;
θ₁ is the initial angle between the dipole moment and the magnetic field
is the final angle which is zero (0) since the dipole moment is aligned with the magnetic field
μB(cos0 - cosθ₁) = 0.458 mJ
Substitute the given values of μ and B
0.0243 x 0.0575 (1 - cosθ₁) = 0.000458
0.00139725 (1 - cosθ₁) = 0.000458
(1 - cosθ₁) = 0.000458 / 0.00139725
(1 - cosθ₁) = 0.327787
cosθ₁ = 1 - 0.327787
cosθ₁ = 0.672213
θ₁ = cos⁻¹ (0.672213)
θ₁ = 47.76⁰
Thus, the initial angle between the dipole moment and the magnetic field is 47.76⁰
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Answer:
a) E = 8.63 10³ N /C, E = 7.49 10³ N/C
b) E= 0 N/C, E = 7.49 10³ N/C
Explanation:
a) For this exercise we can use Gauss's law
Ф = ∫ E. dA = /ε₀
We must take a Gaussian surface in a spherical shape. In this way the line of the electric field and the radi of the sphere are parallel by which the scalar product is reduced to the algebraic product
The area of a sphere is
A = 4π r²
if we use the concept of density
ρ = q_{int} / V
q_{int} = ρ V
the volume of the sphere is
V = 4/3 π r³
we substitute
E 4π r² = ρ (4/3 π r³) /ε₀
E = ρ r / 3ε₀
the density is
ρ = Q / V
V = 4/3 π a³
E = Q 3 / (4π a³) r / 3ε₀
k = 1 / 4π ε₀
E = k Q r / a³
let's calculate
for r = 4.00cm = 0.04m
E = 8.99 10⁹ 3.00 10⁻⁹ 0.04 / 0.05³
E = 8.63 10³ N / c
for r = 6.00 cm
in this case the gaussine surface is outside the sphere, so all the charge is inside
E (4π r²) = Q /ε₀
E = k q / r²
let's calculate
E = 8.99 10⁹ 3 10⁻⁹ / 0.06²
E = 7.49 10³ N/C
b) We repeat in calculation for a conducting sphere.
For r = 4 cm
In this case, all the charge eta on the surface of the sphere, due to the mutual repulsion between the mobile charges, so since there is no charge inside the Gaussian surface, therefore the field is zero.
E = 0
In the case of r = 0.06 m, in this case, all the load is inside the Gaussian surface, therefore the field is
E = k q / r²
E = 7.49 10³ N / C