Question

a 1.0 - cm - radius loop of wire with a resistance of 0.010ω is placed in the center of a solenoid so the loops (vec{a}) is aligned to (vec{b}). the solenoid is 20 cm long and wrapped with 1000 turns of wire. the current through the solenoid increases at a steady rate of 10 a/s. find the current in the loop of wire. mag. field of solenoid (b=mu_{0} n i) emf in a loop (mathcal{e}=-n \frac{d phi}{d t}=-n \frac{d(vec{b} cdot vec{a})}{d t}) (i_{\text {loop }}=\frac{mathcal{e}}{r}=2 mathrm{~ma}) direction of current is determined next slides

Explanation:

Step1: Calculate the magnetic field of the solenoid

The magnetic field of a solenoid is given by \(B = \mu_0nl\), where \(n=\frac{N}{L}\) (number of turns per unit length). Here, \(N = 1000\) turns, \(L=0.2\space m\), so \(n=\frac{1000}{0.2}=5000\space m^{-1}\). The rate of change of current \(\frac{di}{dt}=10\space A/s\).

The rate of change of magnetic field \(\frac{dB}{dt}=\mu_0n\frac{di}{dt}\), with \(\mu_0 = 4\pi\times 10^{-7}\space T\cdot m/A\).

\(\frac{dB}{dt}=4\pi\times 10^{-7}\times5000\times10\)

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Step2: Calculate the induced emf in the loop

The area of the loop \(A=\pi r^{2}\), with \(r = 0.01\space m\), so \(A=\pi\times(0.01)^{2}= \pi\times10^{-4}\space m^{2}\).

The induced emf in the loop (using \(\mathcal{E}=-\frac{d\Phi}{dt}\), and since \(N = 1\) turn for the loop and \(\Phi=BA\)) is \(\mathcal{E}=A\frac{dB}{dt}\)

Substitute \(A=\pi\times 10^{-4}\space m^{2}\) and \(\frac{dB}{dt}=20\pi\times 10^{-3}\space T/s\)

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Step3: Calculate the current in the loop

Using \(I=\frac{\mathcal{E}}{R}\), with \(R = 0.010\space\Omega\)

\(I=\frac{1.972\times 10^{-5}}{0.010}=1.972\times 10^{-3}\space A\approx 2\space mA\)

Answer:

The current in the loop of wire is \(2\space mA\)