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Show that the magnetic field of a dipole can be written in coordinate-free form:

Bdip(r)=μ04π1r3[3(m⋅r^)r^-m]

Short Answer

Expert verified

The magnetic field of a dipoleμ04Ï€°ù3[(3m⃗⋅r^)r^-m⃗] has been proved.

Step by step solution

01

Significance of the magnetism

Magnetism is a type of physical phenomenon produced by a motion of electric charges. Magnetism significantly results in repulsive or attractive force amongst the objects.

02

Determination of the magnetic field of a dipole

The equation of the dipole magnetic field is expressed as:

B=μ0m4Ï€°ù32cosθr^+sinθθ^=μ0m4Ï€°ù3z^ ...... (i)

Here, μ0is the permeability, m is the magnetic dipole moment ,r is the distance between the dipole charges, z^is the position vector in the z direction, and θis the angle between the dipoles.

If the dipole orients towards the z axis, then the equation of the magnetic dipole moment can be expressed as:

m→=mz^

Here, m→is the magnetic dipole moment vector and z^is the position vector in the z direction.

Substitute ³¦´Ç²õθr^-²õ¾±²Ôθθ^forz^in the above equation.

m→=m³¦´Ç²õθr^-²õ¾±²Ôθθ^=m³¦´Ç²õθr^-m²õ¾±²Ôθθ^=m→.r^r^-m³¦´Ç²õθr^-m→=2m→.r^r^+m→.r^r^-m→

Hence, further as:

m→=2m→.r^r^+m→.r^r^-m→=3m→.r^r^-m→

Substitute the above value in equation (i).

localid="1657530579098" B=μ04Ï€°ù3[(3m⃗⋅r^)r^-m⃗]

Thus, the magnetic field of a dipolelocalid="1657528216508" μ04Ï€°ù3[(3m⃗⋅r^)r^-m⃗]has been proved.

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Most popular questions from this chapter

thick slab extending from z=-ato z=+a(and infinite in the x andy directions) carries a uniform volume current J=Jx^(Fig. 5.41). Find the magnetic field, as a function of z, both inside and outside the slab.

The magnetic field on the axis of a circular current loop (Eq. 5.41) is far from uniform (it falls off sharply with increasing z). You can produce a more nearly uniform field by using two such loops a distanced apart (Fig. 5.59).

(a) Find the field (B) as a function of z, and show that ∂B∂z is zero at the point midway between them (z = 0)

(b) If you pick d just right, the second derivative of B will also vanish at the midpoint. This arrangement is known as a Helmholtz coil; it's a convenient way of producing relatively uniform fields in the laboratory. Determine d such that ∂2B/∂z2=0 at the midpoint, and find the resulting magnetic field at the center. [Answer:8μ0I55R ]

Show that the magnetic field of an infinite solenoid runs parallel to the axis, regardless of the cross-sectional shape of the coil,as long as that shape is constant along the length of the solenoid. What is the magnitude of the field, inside and outside of such a coil? Show that the toroid field (Eq. 5.60) reduces to the solenoid field, when the radius of the donut is so large that a segment can be considered essentially straight.

A current flows to the right through a rectangular bar of conducting material, in the presence of a uniform magnetic fieldBpointing out of the page (Fig. 5.56).

(a) If the moving charges are positive, in which direction are they deflected by the magnetic field? This deflection results in an accumulation of charge on the upper and lower surfaces of the bar, which in turn produces an electric force to counteract the magnetic one. Equilibrium occurs when the two exactly cancel. (This phenomenon is known as the Hall effect.)

(b) Find the resulting potential difference (the Hall voltage) between the top and bottom of the bar, in terms ofB,v(the speed of the charges), and the relevant dimensions of the bar.23

(c) How would your analysis change if the moving charges were negative? [The Hall effect is the classic way of determining the sign of the mobile charge carriers in a material.]

Consider a planeloop of wire that carries a steady current I;we

want to calculate the magnetic field at a point in the plane. We might as well take

that point to be the origin (it could be inside or outside the loop). The shape of the

wire is given, in polar coordinates, by a specified function r(θ)(Fig. 5.62).

(a) Show that the magnitude of the field is

role="math" localid="1658927560350" B=μ0I4π∮(5.92)

(b) Test this formula by calculating the field at the center of a circular loop.

(c) The "lituus spiral" is defined by a

r(θ)=aθ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰0<θ≤2Ï€

(for some constant a).Sketch this figure, and complete the loop with a straight

segment along the xaxis. What is the magnetic field at the origin?

(d) For a conic section with focus at the origin,

r(θ)=p1+e³¦´Ç²õθ

where pisthe semi-latus rectum (the y intercept) and eis the eccentricity (e= 0

for a circle, 0 < e< 1 for an ellipse, e= 1 for a parabola). Show that the field is

B=μ0I2pregardless of the eccentricity.

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