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Reika [66]
2 years ago
11

Which of the following will cause an induced current in a coil of wire? Which of the following will cause an induced current in

a coil of wire? a magnet resting near the coil the constant field of the earth passing through the coil a wire carrying a constant current near the coil a magnet being moved into or out of the coil
Physics
1 answer:
svp [43]2 years ago
6 0

Answer:

a magnet being moved into or out of the coil

Explanation:

Electromagnetic induction on a coil of wire depends on relative motion between the wire and a magnetic field. If there is no relative motion between the magnet and the coil, no current is included on the wire. This implies that a stationary magnet around a coil of wire will not produce induced current on a stationary conductor or wire.

Summarily, whenever a conductor is placed in a changing magnetic field, current is induced on the conductor.

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series RC circuit is built with a 15 kΩ resistor and a parallel-plate capacitor with 18-cm-diameter electrodes. A 18 V, 36 kHz s
andre [41]

Answer:

d=1.84\ mm

Explanation:

<u>Capacitance</u>

A two parallel-plate capacitor has a capacitance of

\displaystyle C=\frac{\epsilon_o A}{d}

where

\epsilon_o=8.85\cdot 10^{-12}\ F/m

A = area of the plates = \pi r^2

d = separation of the plates

\displaystyle d=\frac{\epsilon_o A}{C}=\frac{\epsilon_o \pi r^2}{C}

We need to compute C. We'll use the circuit parameters for that. The reactance of a capacitor is given by

\displaystyle X_c=\frac{1}{wC}

where w is the angular frequency

w=2\pi f=2\pi \cdot 36000=226194.67\ rad/s

Solving for C

\displaystyle C=\frac{1}{wX_c}

The reactance can be found knowing the total impedance of the circuit:

Z^2=R^2+X_c^2

Where R is the resistance, R=15 K\Omega=15000\Omega. Solving for Xc

X_c^2=Z^2-R^2

The magnitude of the impedance is computed as the ratio of the rms voltage and rms current

\displaystyle Z=\frac{V}{I}

The rms current is the peak current Ip divided by \sqrt{2}, thus

\displaystyle Z=\frac{\sqrt{2}V}{I_p}

I_p=0.65\ mA/1000=0.00065\ A

Now collect formulas

\displaystyle X_c^2=Z^2-R^2=\left(\frac{\sqrt{2}V}{I_p}\right)^2-R^2

Or, equivalently

\displaystyle X_c=\sqrt{\frac{2V^2}{I_p^2}-R^2}

\displaystyle X_c=\sqrt{\frac{2\cdot 18^2}{0.00065^2}-15000^2}

X_c=36176.34\ \Omega

The capacitance is now

\displaystyle C=\frac{1}{226194.67\cdot 36176.34}=1.22\cdot 10^{-10}\ F

The radius of the plates is

r=18\ cm/2=9 \ cm = 0.09 \ m

The separation between the plates is

\displaystyle d=\frac{8.85\cdot 10^{-12} \cdot \pi\cdot 0.09^2}{1.22\cdot 10^{-10}}

d=0.00184\ m

\boxed{d=1.84\ mm}

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Answer:

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Answer:

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Explanation:

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