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Chapter 29: Problem 52

An emf of \(20.0 \mathrm{~V}\) is applied to a coil with an inductance of \(40.0 \mathrm{mH}\) and a resistance of \(0.500 \Omega\). a) Determine the energy stored in the magnetic field when the current reaches one fourth of its maximum value. b) How long does it take for the current to reach this value?

Short Answer

Expert verified
Answer: The energy stored in the magnetic field when the current reaches one-fourth of its maximum value is 2.00 J, and it takes 27.73 ms for the current to reach that value.

Step by step solution

01

Determine the maximum current

To find the maximum current in the coil, use Ohm's law with the provided emf and resistance: $$I_{max} = \frac{E}{R}$$ Plugging in the given values, we have: $$I_{max} = \frac{20.0\mathrm{~V}}{0.500\Omega} = 40.0\mathrm{~A}$$
02

Find the current as a function of time

Considering the inductance and resistance in the circuit, the current as a function of time is given by: $$I(t) = I_{max}(1 - e^{-\frac{Rt}{L}})$$ Now, we want to find when the current reaches one-fourth of its maximum value: $$I_{\frac{1}{4}} = \frac{1}{4} I_{max}$$
03

Determine the time it takes to reach one-fourth of the maximum current

Equating the above expressions, we have: $$\frac{1}{4} I_{max} = I_{max}(1 - e^{-\frac{Rt}{L}})$$ Dividing both sides by \(I_{max}\) and rearranging, we obtain: $$e^{-\frac{Rt}{L}} = 1 - \frac{1}{4} \Rightarrow -\frac{Rt}{L} = \ln\left(\frac{3}{4}\right)$$ Solving for \(t\), we have: $$t = -\frac{L}{R}\ln\left(\frac{3}{4}\right)$$ Plugging in the given values for \(R\) and \(L\), we find: $$t = -\frac{40.0\mathrm{~mH}}{0.500\Omega}\ln\left(\frac{3}{4}\right) = 27.73\mathrm{~ms}$$
04

Calculate the energy stored in the magnetic field

The energy stored in the magnetic field when the current reaches one-fourth of its maximum value, is given by: $$W = \frac{1}{2}LI_{\frac{1}{4}}^2$$ Using the calculated values from steps 1 and 3, we have: $$W = \frac{1}{2}(40.0\mathrm{~mH})(10.0\mathrm{~A})^2 = 2.00\mathrm{~J}$$ So the energy stored in the magnetic field when the current reaches one-fourth of its maximum value is \(2.00\mathrm{~J}\), and it takes \(27.73\mathrm{~ms}\) for the current to reach that value.

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Most popular questions from this chapter

A \(100 .-V\) battery is connected in series with a \(500 .-\Omega\) resistor. According to Faraday's Law of Induction, current can never change instantaneously, so there is always some "stray" inductance. Suppose the stray inductance is \(0.200 \mu \mathrm{H}\). How long will it take the current to build up to within \(0.500 \%\) of its final value of \(0.200 \mathrm{~A}\) after the resistor is con- nected to the battery?

A wire of length \(\ell=10.0 \mathrm{~cm}\) is moving with constant velocity in the \(x y\) -plane; the wire is parallel to the \(y\) -axis and moving along the \(x\) -axis. If a magnetic field of magnitude \(1.00 \mathrm{~T}\) is pointing along the positive \(z\) -axis, what must the velocity of the wire be in order to induce a potential difference of \(2.00 \mathrm{~V}\) across it?

A metal hoop is laid flat on the ground. A magnetic field that is directed upward, out of the ground, is increasing in magnitude. As you look down on the hoop from above, what is the direction of the induced current in the hoop?

A supersonic aircraft with a wingspan of \(10.0 \mathrm{~m}\) is flying over the north magnetic pole (in a magnetic field of magnitude 0.500 G perpendicular to the ground) at a speed of three times the speed of sound (Mach 3). What is the potential difference between the tips of the wings? Assume that the wings are made of aluminum.

A steel cylinder with radius \(2.5 \mathrm{~cm}\) and length \(10.0 \mathrm{~cm}\) rolls without slipping down a ramp that is inclined at \(15^{\circ}\) above the horizontal and has a length (along the ramp) of \(3.0 \mathrm{~m} .\) What is the induced potential difference between the ends of the cylinder at the bottom of the ramp, if the surface of the ramp points along magnetic north?
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