A round flat metal coil has 180 turns and negligible resistance. It is connected in a series circuit with a $17 - \Omega$ resistor, with nothing else in the circuit. You measure that a $6.0 - \mathrm { A }$ current flows through the resistor when a magnetic field through the coil, perpendicular to its area, is changing at $3.0 \mathrm {~T} / \mathrm { s }$ What is the radius of the coil?

Question

Quizplus · Accepted Answer

The magnetic flux through the coil is given by $\Phi = BA$ where $B$ is the magnetic field and $A$ is the area of the coil. The emf induced in the coil is given by $\varepsilon = -d\Phi/dt$. The current in the circuit is given by $\varepsilon = IR$, where $I$ is the current and $R$ is the resistance. Combining these equations, we get $I = -A(dB/dt)/R$. Solving for $A$, we get $A = IR/(dB/dt)$. Substituting the given values, we get $A = (6.0 \mathrm { A})(17 \Omega)/(3.0 \mathrm { T} / \mathrm { s }) = 34 \mathrm { m }^2$. The area of a circle is given by $A = \pi r^2$, so the radius of the coil is $r = \sqrt{A/\pi} = \sqrt{34 \mathrm { m }^2/\pi} = 0.25 \mathrm { m }$.

A Round Flat Metal Coil Has 180 Turns and Negligible 17−Ω17 - \Omega17−Ω

A Round Flat Metal Coil Has 180 Turns and Negligible $17 - \Omega$