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Chapter 29: Problem 22

A circular coil of wire with 20 turns and a radius of \(40.0 \mathrm{~cm}\) is laying flat on a horizontal table as shown in the figure. There is a uniform magnetic field extending over the entire table with a magnitude of \(5.00 \mathrm{~T}\) and directed to the north and downward, making an angle of \(25.8^{\circ}\) with the horizontal. What is the magnitude of the magnetic flux through the coil?

Short Answer

Expert verified
Question: Calculate the magnitude of the magnetic flux through a circular coil of 20 turns, radius 40 cm, and placed in a uniform magnetic field of 5.00 T, directed at an angle of 25.8° downward from the horizontal. Answer: First, calculate the area of the circular coil: \(A = \pi (0.4)^2\) \(A \approx 0.5027 \, \mathrm{m}^2\) Next, find the angle between the magnetic field and the normal to the coil: \(\theta = 90^{\circ} - 25.8^{\circ}\) \(\theta = 64.2^{\circ}\) Finally, calculate the magnetic flux through the coil: \(\Phi_B = (20)(5.00)(0.5027)(\cos(64.2^{\circ}))\) \(\Phi_B \approx 50.27 \, \mathrm{Wb}\) The magnitude of the magnetic flux through the circular coil is approximately 50.27 Wb.

Step by step solution

01

Calculate the Area of the Circular Coil

The area of a circle can be calculated using the formula: \(A = \pi r^2\) Where, A = Area r = radius Given, radius r = 40.0 cm = 0.4 m (as we need to convert it to meters) Therefore, the area A of the circular coil will be: \(A = \pi (0.4)^2\)
02

Calculate the angle between the magnetic field and the normal to the coil

The magnetic field is making an angle of \(25.8^{\circ}\) downward from the horizontal. Thus, we need to find the angle between the magnetic field and the vertical axis (normal to the coil) which is: \(\theta = 90^{\circ} - 25.8^{\circ}\)
03

Calculate the Magnetic Flux through the coil

Now, we have all the required values to calculate the magnetic flux through the circular coil. Using the formula: \(\Phi_B = NBA\cos\theta\) Where, N = 20 turns B = 5.00 T A = \(\pi (0.4)^2 \,\mathrm{m}^2\) \(\theta = (90^{\circ} - 25.8^{\circ})\) We can now plug in all the values and calculate the magnetic flux: \(\Phi_B = (20)(5.00)(\pi (0.4)^2)(\cos(90^{\circ} - 25.8^{\circ}))\) Calculate the value of magnetic flux to get the magnitude of the magnetic flux through the coil.

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Most popular questions from this chapter

Two parallel conducting rails with negligible resistance are connected at one end by a resistor of resistance \(R\), as shown in the figure. The rails are placed in a magnetic field \(\vec{B}_{\text {ext }},\) which is perpendicular to the plane of the rails. This magnetic field is uniform and time independent. The distance between the rails is \(\ell\). A conducting rod slides without friction on top of the two rails at constant velocity \(\vec{v}\). a) Using Faraday's Law of Induction, calculate the magnitude of the potential difference induced in the moving rod. b) Calculate the magnitude of the induced current in the \(\operatorname{rod}, i_{\text {ind }}\). c) Show that for the rod to move at a constant velocity as shown, it must be pulled with an external force, \(\vec{F}_{\mathrm{ext}},\) and calculate the magnitude of this force. d) Calculate the work done, \(W_{\text {ext }},\) and the power generated, \(P_{\text {ext }}\), by the external force in moving the rod. e) Calculate the power used (dissipated) by the resistor, \(P_{\mathrm{R}}\). Explain the correlation between this result and those of part (d).

A short coil with radius \(R=10.0 \mathrm{~cm}\) contains \(N=30.0\) turns and surrounds a long solenoid with radius \(r=8.00 \mathrm{~cm}\) containing \(n=60\) turns per centimeter. The current in the short coil is increased at a constant rate from zero to \(i=2.00 \mathrm{~A}\) in a time of \(t=12.0 \mathrm{~s}\). Calculate the induced potential difference in the long solenoid while the current is increasing in the short coil.

Which of the following will induce a current in a loop of wire in a uniform magnetic field? a) decreasing the strength of the field b) rotating the loop about an axis parallel to the field c) moving the loop within the field d) all of the above e) none of the above

Large electric fields are certainly a hazard to the human body, as they can produce dangerous currents, but what about large magnetic fields? A man \(1.80 \mathrm{~m}\) tall walks at \(2.00 \mathrm{~m} / \mathrm{s}\) perpendicular to a horizontal magnetic field of \(5.0 \mathrm{~T} ;\) that is, he walks between the pole faces of a very big magnet. (Such a magnet can, for example, be found in the National Superconducting Cyclotron Laboratory at Michigan State University.) Given that his body is full of conducting fluids, estimate the potential difference induced between his head and feet.

When you plug a refrigerator into a wall socket, on occasion, a spark appears between the prongs. What causes this?
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