Ring current

A metal ring with a radius of 10 cm is placed in a magnetic field so that the normal to its surface makes an angle of 60° with the induction lines. The induction of the magnetic field varies uniformly and drops from 0.45 T to 0.35 T in 0.50 s. Determine the current in the ring when its resistance is 0.50 Ω.

Final Answer:

I =  -0.0063 A

Step-by-step explanation:

$$\begin{gathered} r=10\text{cm}\to \text{m}=10:100\text{m}=0.1\text{m} \\ \alpha =60{}^{\circ } \\ t=0.50\text{s} \\ {B}_1=0.35\text{T} \\ {B}_2=0.45\text{T} \\ \mu =4\pi \cdot 10^{-7}=4\cdot 3.1416\cdot 10^{-7}\doteq 1.2566\cdot 10^{-6}\text{H/m} \\ S=\pi \cdot r^2=3.1416\cdot 0.1^2\doteq 0.0314{\text{m}}^2 \\ R=0.50\Omega \\ {\Phi }_1=B_1\cdot S\cdot \cos \alpha =B_1\cdot S\cdot \cos 60°=0.35\cdot 0.0314\cdot \cos 60°=0.35\cdot 0.0314\cdot 0.5=0.0055=7\pi /4000\text{Wb} \\ {\Phi }_2=B_2\cdot S\cdot \cos \alpha =B_2\cdot S\cdot \cos 60°=0.45\cdot 0.0314\cdot \cos 60°=0.45\cdot 0.0314\cdot 0.5=0.00707=9\pi /4000\text{Wb} \\ U=\frac{{\Phi }_1-{\Phi }_2}{t}=\frac{0.0055-0.0071}{0.5}\doteq -0.0031\text{V} \\ I=U/R=(-0.0031)/0.5=-0.0063\text{A} \end{gathered}$$

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