Question

A washer is made of metal of having resistivity ${{10}^{-7}}\Omega m$. The washer has inner radius 1 cm, outer radius 3 cm and thickness 1 mm. A magnetic field, oriental to the plane of the washer, has the time dependent magnitude $B=(2t)T{{s}^{-1}}$. Find the current (in amperes) around the washer.
A) 2 A
B) 4 A
C) 6 A
D) 8 A

Answer 1

We need to understand how to develop a useful relationship between the resistivity of a material, the magnetic field through the plane of the material and its dimensions as given in the problem to solve for the current through the metal.

Complete answer:
We are given a metallic washer, which is in the shape of a ring with the inner radius as 1 cm and the outer radius as 3 cm. We can find the area of the ring made up the metal as –

& {{A}_{eff}}=\pi {{r}_{out}}^{2}-\pi {{r}_{in}}^{2} \\\ & \Rightarrow {{A}_{eff}}=\pi ({{3}^{2}}-{{1}^{2}}) \\\ & \Rightarrow {{A}_{eff}}=8\times {{10}^{-4}}\pi {{m}^{2}} \\\ \end{aligned}$$  The cross-sectional area of the metallic washer can be given as – $$\begin{aligned} & {{A}_{cross}}=({{r}_{out}}-{{r}_{in}}).\text{thickness} \\\ & \Rightarrow {{A}_{cross}}=2cm.1mm \\\ & \Rightarrow {{A}_{cross}}=2\times {{10}^{-5}}{{m}^{2}} \\\ \end{aligned}$$ Now, we know the resistivity of the given metal as $${{10}^{-7}}\Omega m$$. We know that the resistance is related to the resistivity with the area of the metal and the length of the metal wire. The length of the metal wire can be found using the average radius between the outer and the inner radius as – $$\begin{aligned} & l=2\pi {{r}_{avg}} \\\ & \Rightarrow l=2\pi ({{r}_{out}}-{{r}_{in}}) \\\ & \Rightarrow l=2\pi (3-1) \\\ & \Rightarrow l=4\pi \times {{10}^{-2}}m \\\ \end{aligned}$$ Now, we can find the resistance of the metal washer as – $$R=\dfrac{\rho l}{{{A}_{eff}}}$$ Now, we know from Faraday's law of electromagnetic induction that the rate of change flux gives the electromotive force (emf) induced in the metal. We can write it as- $$\begin{aligned} & \varepsilon =\dfrac{d\phi }{dt} \\\ & \text{Also,} \\\ & \phi =B.{{A}_{cross}} \\\ & \Rightarrow \phi =2t.{{A}_{cross}} \\\ & \Rightarrow \varepsilon =\dfrac{d}{dt}(2t).{{A}_{cross}} \\\ & \Rightarrow \varepsilon =2.{{A}_{cross}} \\\ \end{aligned}$$ Now, we can find the current through the metallic washer using the Ohm’s law as – $$\begin{aligned} & V=IR \\\ & \Rightarrow I=\dfrac{V}{R} \\\ & \Rightarrow I=\dfrac{2{{A}_{cross}}}{\dfrac{\rho l}{{{A}_{eff}}}} \\\ & \Rightarrow I=\dfrac{2{{A}_{cross}}{{A}_{eff}}}{\rho l} \\\ & \Rightarrow I=\dfrac{2(2\times {{10}^{-5}})(8\pi \times {{10}^{-4}})}{{{10}^{-7}}\times 4\pi \times {{10}^{-2}}m} \\\ & \therefore I=8A \\\ \end{aligned}$$ So, the current through the washer is 8A. **The correct answer is option D.** **Note:** The area of the cross-section of the metal washer is considered as a rectangular element of thickness 1 mm. The area of the washer has to be found by subtracting the area of the inner radius from that of the outer radius and not after finding the average radius of the washer.