Physics

Physics · electromagnetism

Problem

c. Derive an expression for the magnetic field B→i\overrightarrow{\mathbf{B}}_{i} from the iron a distance d≫ad \gg a from the center of the ball. Note that there will be a component directed radially away from the ball and a component directed tangent to a circle of radius dd around the ball, so using polar coordinates is recommended. Given constants: g=9.8 N/kgg=9.8 \mathrm{~N} / \mathrm{kg} k=1/4πϵ0=8.99×109 N⋅m2/C2k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} c=3.00×108 m/sc=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} NA=6.02×1023( mol)−1N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} σ=5.67×10−8 J/(s⋅m2⋅K4)\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) 1eV=1.602×10−19 J1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} me=9.109×10−31 kg=0.511MeV/c2m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} mp=1.673×10−27 kg=938MeV/c2m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2} sin⁡θ≈θ−16θ3\sin \theta \approx \theta-\frac{1}{6} \theta^{3} for ∣θ∣≪1|\theta| \ll 1 G=6.67×10−11 N⋅m2/kg2km=μ0/4π=10−7 T⋅m/AkB=1.38×10−23 J/KR=NAkB=8.31 J/(mol⋅K)e=1.602×10−19Ch=6.63×10−34 J⋅s=4.14×10−15eV⋅s(1+x)n≈1+nx for ∣x∣≪1ln⁡(1+x)≈x for ∣x∣≪1cos⁡θ≈1−12θ2 for ∣θ∣≪1\begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned}  velocity parameter β=vc Lorentz factor γ=11−β2 relativistic momentum p=γβmc relativistic energy E=γmc2 relativistic doppler shift ff0=1−β1+β\begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} Extra Supplementary Reading Materials: A ship can be thought of as a symmetric arrangement of soft iron. In the presence of an external magnetic field, the soft iron will become magnetized, creating a second, weaker magnetic field. We want to examine the effect of the ship's field on the ship's compass, which will be located in the middle of the ship. Let the strength of the Earth's magnetic field near the ship be BeB_{e}, and the orientation of the field be horizontal, pointing directly toward true north. The Earth's magnetic field BeB_{e} will magnetize the ship, which will then create a second magnetic field BsB_{s} in the vicinity of the ship's compass given by B→s=Be(−Kbcos⁡θb^+Kssin⁡θs^)\overrightarrow{\mathbf{B}}_{s}=B_{e}\left(-K_{b} \cos \theta \hat{\mathbf{b}}+K_{s} \sin \theta \hat{\mathbf{s}}\right) where KbK_{b} and KsK_{s} are positive constants, θ\theta is the angle between the heading of the ship and magnetic north, measured clockwise, b^\hat{\mathbf{b}} and s^\hat{\mathbf{s}} are unit vectors pointing in the forward direction of the ship (bow) and directly right of the forward direction (starboard), respectively. Because of the ship's magnetic field, the ship's compass will no longer necessarily point North. Context question: a. Derive an expression for the deviation of the compass, δθ\delta \theta, from north as a function of KbK_{b}, KsK_{s}, and θ\theta. Context answer: δθ=arctan⁡(Ks−Kb)sin⁡θcos⁡θ1−Kbcos⁡2θ−Kssin⁡2θ\delta \theta=\arctan \left(K_{s}-K_{b}\right) \frac{\sin \theta \cos \theta}{1-K_{b} \cos ^{2} \theta-K_{s} \sin ^{2} \theta} Context question: b. Assuming that KbK_{b} and KsK_{s} are both much smaller than one, at what heading(s) θ\theta will the deviation δθ\delta \theta be largest? Context answer: 4545, 135135, 225225, 315315 Extra Supplementary Reading Materials: A pair of iron balls placed in the same horizontal plane as the compass but a distance dd away can be used to help correct for the error caused by the induced magnetism of the ship. <image_1> A binnacle, protecting the ship's compass in the center, with two soft iron spheres to help correct for errors in the compass heading. The use of the spheres was suggested by Lord Kelvin. Just like the ship, the iron balls will become magnetic because of the Earth's field BeB_{e}. As spheres, the balls will individually act like dipoles. A dipole can be thought of as the field produced by two magnetic monopoles of strength ±m\pm m at two different points. The magnetic field of a single pole is B→=±mr^r2\overrightarrow{\mathbf{B}}= \pm m \frac{\hat{\mathbf{r}}}{r^{2}} where the positive sign is for a north pole and the negative for a south pole. The dipole magnetic field is the sum of the two fields: a north pole at y=+a/2y=+a / 2 and a south pole at y=−a/2y=-a / 2, where the yy axis is horizontal and pointing north. aa is a small distance much smaller than the radius of the iron balls; in general a=KiBea=K_{i} B_{e} where KiK_{i} is a constant that depends on the size of the iron sphere. <image_2>

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