Mastering Physics Solutions: Electric and Magnetic Field Vectors Conceptual Question

Mastering Physics Solutions: Electric and Magnetic Field Vectors Conceptual Question

On May 19, 2012, in Chapter 20: Electromagnetic Induction and Waves, by Mastering Physics Solutions

Part A = +z
Part B = -z
Part C = +y
Part D = at a –45° angle in the xz plane

The electric and magnetic field vectors at a specific point in space and time are illustrated. Based on this information, in what direction does the electromagnetic wave propagate?

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Mastering Physics Solutions: Exercise 19.14

Mastering Physics Solutions: Exercise 19.14

On May 10, 2012, in Chapter 19: Magnetism, by Mastering Physics Solutions

Part A = 75000 m/s Click to use the calculator/solver for this part of the problem
Part B = 75000 Click to use the calculator/solver for this part of the problem

A charged particle travels undeflected through perpendicular electric and magnetic fields whose magnitudes are 3000 N/C and 40 mT, respectively.
Find the speed of the particle if it is a proton.
Find the speed of the particle if it is an alpha particle. (An alpha particle is a helium nucleus-a positive ion with a double positive charge of +2e.)

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Mastering Physics Solutions: A Conductor Moving in a Magnetic Field

Mastering Physics Solutions: A Conductor Moving in a Magnetic Field

On May 7, 2012, in Chapter 19: Magnetism, by Mastering Physics Solutions

Part A = 5 N/C
Part B = -x̂
Part C = -x̂

A metal cube with sides of length a = 1 cm is moving at velocity v = 1 m/s ŷ across a uniform magnetic field B = 5 T ẑ. The cube is oriented so that four of its edges are parallel to its direction of motion (i.e., the normal vectors of two faces are parallel to the direction of motion).
Find E, the magnitude of the electric field inside the cube.
What is the direction of the induced electric field?
If one replaces the conducting cube with one that has positive charge carriers, what is the direction of the induced electric field?

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Mastering Physics Solutions: Exercise 16.39

Mastering Physics Solutions: Exercise 16.39

On February 15, 2012, in Chapter 16: Electric Potential, Energy, and Capacitance, by Mastering Physics Solutions

Part A = 6.31 * 10-9 C Click to use the calculator/solver for this part of the problem
Part B = 3.79 * 10-8 J Click to use the calculator/solver for this part of the problem
Part B = 2700 V/m Click to use the calculator/solver for this part of the problem

A 12.0 V battery remains connected to a parallel plate capacitor with a plate area of 0.264 m2 and a plate separation of 4.44 mm.
What is the charge on the capacitor?
How much energy is stored in the capacitor?
What is the electric field between its plates?

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Mastering Physics Solutions: Energy in Capacitors and Electric Fields

Mastering Physics Solutions: Energy in Capacitors and Electric Fields

On February 12, 2012, in Chapter 16: Electric Potential, Energy, and Capacitance, by Mastering Physics Solutions

Part A = U = Q2 / (2C)
Part B = U = (CV2) / 2
Part C = U/U0 = 0.5
Part D = U/U0 = 2
Part E = U = (ε0AV2) / 2d
Part F = U = (ε0AdE2) / 2
Part G = u = (ε0E2) / 2

Find the energy U of the capacitor in terms of C and Q by using the definition of capacitance and the formula for the energy in a capacitor.
Find the energy U of the capacitor in terms of C and V by using the definition of capacitance and the formula for the energy in a capacitor.
A parallel-plate capacitor is connected to a battery. The energy of the capacitor is U0. The capacitor remains connected to the battery while the plates are slowly pulled apart until the plate separation doubles. The new energy of the capacitor is U. Find the ratio U/U0.
A parallel-plate capacitor is connected to a battery. The energy of the capacitor is U0. The capacitor is then disconnected from the battery and the plates are slowly pulled apart until the plate separation doubles. The new energy of the capacitor is U. Find the ratio U/U0.
A parallel-plate capacitor has area A and plate separation d, and it is charged to voltage V. Use the formulas from the problem introduction to obtain the formula for the energy U of the capacitor.
A parallel-plate capacitor has area A and plate separation d, and it is charged so that the electric field inside is E. Use the formulas from the problem introduction to find the energy U of the capacitor.
Find the energy density u of the electric field in a parallel-plate capacitor. The magnitude of the electric field inside the capacitor is E.

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Mastering Physics Solutions: Electric Fields and Equipotential Surfaces

Mastering Physics Solutions: Electric Fields and Equipotential Surfaces

On February 6, 2012, in Chapter 16: Electric Potential, Energy, and Capacitance, by Mastering Physics Solutions

Part A = 0 J
Part B = 1 J
Part C = greater than the magnitude of the electric field at point B.

What is the work WAB done by the electric force to move a 1 C charge from A to B?
What is the work WAD done by the electric force to move a 1 C charge from A to D?
The magnitude of the electric field at point C is

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