Magnetism Exercise (28-08-2020)

Consider a long current carrying cylindrical conductor of radius r r
Current density j j inside the conductor is uniform over its cross-section.
Deduce suitable expression for force of interaction per unit length between two halves that are obtained by dividing the conductor by a plane containing the axis of the conductor.

Answer comes in the form of F = α μ 0 j β r γ F=\alpha \mu_{0}j^{\beta}r^{\gamma} Type your answer as α + β + γ = ? \alpha+\beta+\gamma= ?

The problem is taken from my Physics Book.


The answer is 5.334.

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1 solution

Steven Chase
Aug 27, 2020

Nice problem. The infinitesimal current element is:

d I = ( 0 , 0 , j r d r d θ ) \vec{dI} = (0,0,j r \, dr \, d\theta)

The scalar magnetic flux density at radius r r is (by convention, I consider R R to be the total radius):.

B = μ 0 I e n c l o s e d 2 π r = μ 0 2 π r j π r 2 = μ 0 r j 2 B = \frac{\mu_0 I_{enclosed}}{2 \pi r} = \frac{\mu_0}{2 \pi r} j \pi r^2 = \frac{\mu_0 r j }{2}

The vector magnetic flux density is:

B = ( μ 0 r j 2 sin θ , μ 0 r j 2 cos θ , 0 ) \vec{B} = (- \frac{\mu_0 r j }{2} \sin \theta, \frac{\mu_0 r j }{2} \cos \theta, 0 )

The infinitesimal flux contribution per unit length is:

d F = d I × B = ( μ 0 r 2 j 2 2 cos θ d r d θ , μ 0 r 2 j 2 2 sin θ d r d θ , 0 ) \vec{d F} = \vec{d I } \times \vec{B} = ( -\frac{\mu_0 r^2 j^2 }{2} \cos \theta \, dr \, d \theta, \frac{\mu_0 r^2 j^2 }{2} \sin \theta \, dr \, d \theta, 0)

Only the y y force evaluates to a non-zero net value value over the range ( 0 θ π ) (0 \leq \theta \leq \pi) . Evaluating the y y force over the half circle ( y 0 ) (y \geq 0) results in:

F y = 0 π 0 R μ 0 r 2 j 2 2 sin θ d r d θ = 1 3 μ 0 j 2 R 3 F_y = \int_0^\pi \int_0^R \frac{\mu_0 r^2 j^2 }{2} \sin \theta \, dr \, d \theta = \frac{1}{3} \mu_0 j^2 R^3

@Steven Chase Thanks Upvoted have been awarded.
I only didn't understand that vector magnetic flux density, how did you write that?

Talulah Riley - 9 months, 2 weeks ago

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The radial vector is ( cos θ , sin θ ) (\cos \theta, \sin \theta) . The tangential vector is therefore ( sin θ , cos θ ) (-\sin \theta, \cos \theta) , and the vector magnetic flux density is in that direction.

Steven Chase - 9 months, 2 weeks ago

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@Steven Chase Oh yes, Thanks, smart guy.

Talulah Riley - 9 months, 2 weeks ago

Can you double-check your answer to Magnetism Series #1?

Steven Chase - 9 months, 2 weeks ago

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@Steven Chase The problem is reposted .

Talulah Riley - 9 months, 2 weeks ago

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@Talulah Riley @Steven Chase The problem is again reposted.

Talulah Riley - 9 months, 2 weeks ago

@Steven Chase I didn't understand how to start your new problem?
Can you please give me few starting steps?

Talulah Riley - 9 months, 2 weeks ago

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Find the voltages across the capacitors

Steven Chase - 9 months, 2 weeks ago

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