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wlad13 [49]
3 years ago
13

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

.5 mT. The rotation of the dipole due to the magnetic force on it is unimpeded. When the dipole rotates through the orientation where its dipole moment is aligned with the magnetic field, its kinetic energy is 0.458 mJ. (a) What is the initial angle between the dipole moment and the magnetic field
Physics
2 answers:
ololo11 [35]3 years ago
7 0

Answer:

47.76°

Explanation:

Magnitude of dipole moment = 0.0243J/T

Magnetic Field = 57.5mT

kinetic energy = 0.458mJ

∇U = -∇K

Uf - Ui = -0.458mJ

Ui - Uf = 0.458mJ

(-μBcosθi) - (-μBcosθf) = 0.458mJ

rearranging the equation,

(μBcosθf) - (μBcosθi) = 0.458mJ

μB * (cosθf - cosθi) = 0.458mJ

θf is at 0° because the dipole moment is aligned with the magnetic field.

μB * (cos 0 - cos θi) = 0.458mJ

but cos 0 = 1

(0.0243 * 0.0575) (1 - cos θi) = 0.458*10⁻³

1 - cos θi = 0.458*10⁻³ / 1.397*10⁻³

1 - cos θi = 0.3278

collect like terms

cosθi = 0.6722

θ = cos⁻ 0.6722

θ = 47.76°

Valentin [98]3 years ago
3 0

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

(-\mu Bcos \theta_i )- (-\mu Bcos \theta_f) = 0.458 mJ\\\\-\mu Bcos \theta_i  + \mu Bcos \theta_f = 0.458 mJ\\\\\mu Bcos \theta_f  -\mu Bcos \theta_i  = 0.458 mJ\\\\\mu B(cos \theta_f - cos \theta_i ) = 0.458 mJ

where;

θ₁ is the initial angle between the dipole moment and the magnetic field

\theta_f 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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When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

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F=m\frac{v^2}{r}

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qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

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The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

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E)

For this part, we use again the relationship found in part B:

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6 0
3 years ago
In need help I need someone that is really good at physics
Aloiza [94]
10/70×360°
=51.4°

hope thus helps
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3 years ago
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