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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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