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Vladimir79 [104]
2 years ago
13

A square loop of length 0.1 m on each side is sitting in a magnetic field that is increasing at a rate of 0.1 T/second. What can

we say about the induced EMF?
Physics
1 answer:
Zolol [24]2 years ago
4 0

Answer:

The induced emf can be found by Faraday’s Law.

\epsilon = -\frac{d\Phi_B}{dt}

\Phi_B = BA = B(0.1)^2 = 0.01B

The magnetic field is increasing at a rate of 0.1T/s. So,

\frac{dB}{dt} = 0.1

Finally,

\epsilon = \frac{dBA}{dt} = 0.01\frac{dB}{dt} = 0.01\times 0.1 = 0.001 V

Explanation:

Faraday’s Law states that a change in the magnetic flux induces an emf in the circuit. The magnetic flux is the multiplication of magnetic field and the area of the loop. The area of the loop is simple, and the change of magnetic field as a function of time is given in the question.

The minus sign in front of the Faraday’s Law means that the induced current always opposes the change of the magnetic flux. Since we do not know the direction of the magnetic field in this question, we cannot find the direction of the induced emf on the loop.

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A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo
-BARSIC- [3]

Answer:

It would take \tau(\ln 9 - \ln 8) time for the capacitor to discharge from q_0 to \displaystyle \frac{8}{9} \, q_0.

It would take \tau(\ln 9 - \ln 7) time for the capacitor to discharge from q_0 to \displaystyle \frac{7}{9}\, q_0.

Note that \ln 9 = 2\,\ln 3, and that\ln 8 = 3\, \ln 2.

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is \tau, and the initial charge of the capacitor be q_0. Then at time t, the charge stored in the capacitor would be:

\displaystyle q(t) = q_0 \, e^{-t / \tau}.

<h3>a)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{8}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}.

\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9.

t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8).

<h3>b)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{7}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}.

\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9.

t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7).

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