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Question: Use Eq. 5.41 to obtain the magnetic field on the axis of the rotating disk in Prob. 5.37(a). Show that the dipole field (Eq. 5.88), with the dipole moment you found in Prob. 5.37, is a good approximation if z>> R.

Short Answer

Expert verified

The magnetic field on the axis of the disk is Bd=μ0σӬR48z3. The dipole field with dipole moment is good approximation for very small distance from centre of rotating disk to find magnetic field on the axis of the rotating disk.

Step by step solution

01

Determine the magnetic field on the axis of rotating disk

The total charge on the small element of ring is given as:

»å²Ï2à¤Ê± (dr)

Here, σis the surface charge density for the rotating disk.

The time period for the revolution of disk is given as:

dt=ϑӬ

Here, Ó¬is the angular velocity of rotating disk.

The current in the small element of the disk is given as:

\begin{aligned}&I=\frac{dQ}{dt}\\&I=\frac{\sigma(2\pir)dr}{\left(\frac{2\pi}{\omega}\right)}\\&I=\sigma(r\omega)dr\end{aligned}

The magnetic field on the axis of rotating disk due to small element of disk is given as:

\begin{aligned}&dB=\frac{\mu_{0}l}{2}\left(\frac{r^{2}}{\left(r^{2}+z^{2}\right)^{3/2}}\right)\\&dB=\frac{\mu_{0}(\sigma(r\omega)dr)}{2}\left(\frac{r^{2}}{\left(r^{2}+z^{2}\right)^{3/2}}\right)\end{aligned}

The total magnetic field on the axis of rotating disk is given as:

\begin{aligned}B&=\int_{0}^{R}dB\\B&=\int_{0}^{R}\left(\frac{\mu_{0}(\sigma(r\omega)dr)}{2}\left(\frac{r^{2}}{\left(r^{2}+z^{2}\right)^{3/2}}\right)\right)\\B&=\frac{\mu_{0}\sigma\omega}{2}\left(\frac{R^{2}+2z^{2}}{\sqrt{R^{2}+z^{2}}}-2z\right)\end{aligned}

Apply the approximation z>>Rin the above expression.

B=μ0σӬ2R2+2z2R2+z2-2z

\begin{aligned}&B=\frac{\mu_{0}\sigma\omega}{2}\left[2z\left(1+\frac{R^{2}}{2z^{2}}\right)\left(1-\frac{R^{2}}{2z^{2}}+\frac{3}{8}\left(\frac{R^{4}}{z^{4}}\right)\right)-1\right]\\&B=\frac{\mu_{0}\sigma\omegaR^{4}}{2z^{3}}\end{aligned}

02

Determine the dipole field with approximation

The dipole moment for the rotating disk by equation 5.37 is given as:

m=πσӬR44

The dipole field for the rotating disk is given as:

Bd=μ0m4πr3(2cosθ+sinθ)

The points on the zaxis z=rand θ=0.

Substitute all the values in the above equation.

\begin{aligned}&B_{d}=\frac{\mu_{0}\left(\frac{\pi\sigma\omegaR^{4}}{4}\right)}{4\pi(z)^{3}}(2\cos(0)+\sin(0))\\&B_{d}=\frac{\mu_{0}\sigma\omegaR^{4}}{8z^{3}}\end{aligned}

Therefore, it is clear that to obtain magnetic field on the axis of rotating disk the approximation in dipole field for very small distance compared to radius of disk.

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Most popular questions from this chapter

Question: (a) Find the force on a square loop placed as shown in Fig. 5.24(a), near an infinite straight wire. Both the loop and the wire carry a steady current I.

(b) Find the force on the triangular loop in Fig. 5.24(b).

Suppose you have two infinite straight line chargesλ, a distance d apart, moving along at a constant speed υ(Fig. 5.26). How great would have tobe in order for the magnetic attraction to balance the electrical repulsion? Work out the actual number. Is this a reasonable sort of speed?

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.]

A uniformly charged solid sphere of radius R carries a total charge Q, and is set spinning with angular velocity w about the z axis.

(a) What is the magnetic dipole moment of the sphere?

(b) Find the average magnetic field within the sphere (see Prob. 5.59).

(c) Find the approximate vector potential at a point (r, B) where r>> R.

(d) Find the exact potential at a point (r, B) outside the sphere, and check that it is consistent with (c). [Hint: refer to Ex. 5.11.]

(e) Find the magnetic field at a point (r, B) inside the sphere (Prob. 5.30), and check that it is consistent with (b).

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 ]

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