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: Exercise 19.9

Mastering Physics Solutions: Exercise 19.9

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

Part A = 2.2*10^-18 N Click to use the calculator/solver for this part of the problem
Part B = 1.6 * 10^-18 Click to use the calculator/solver for this part of the problem
Part C = 0 N
Part D = 0 N

An electron travels at a speed of 1.0*10^4 m/s through a uniform magnetic field whose magnitude is 1.4*10^−3 T.
What is the magnitude of the magnetic force on the electron if its velocity and the magnetic field are perpendicular?
What is the magnitude of the magnetic force on the electron if its velocity and the magnetic field make an angle of 45 °?
What is the magnitude of the magnetic force on the electron if its velocity and the magnetic field are parallel?
What is the magnitude of the magnetic force on the electron if its velocity and the magnetic field are exactly opposite?

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Mastering Physics Solutions: Magnetic Materials

Mastering Physics Solutions: Magnetic Materials

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

Part A = paramagnetism
Part B = ferromagnetic
Part C = diamagnetism

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Mastering Physics Solutions: Direction of the Magnetic Field due to a Wire Conceptual Question

Mastering Physics Solutions: Direction of the Magnetic Field due to a Wire Conceptual Question

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

Part A = BA is out of the page.
Part B = BB is into the page.
Part C = BC is out of the page.
Part D = BD is out of the page.
Part E = BE is into the page.

Find the direction of the magnetic field at each of the indicated points.
What is the direction of the magnetic field BA at Point A?
What is the direction of the magnetic field BB at Point B?
What is the direction of the magnetic field BC at Point C?
What is the direction of the magnetic field BD at Point D?
What is the direction of the magnetic field BE at Point E?

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Mastering Physics Solutions: Magnetic Field due to a Wire Conceptual Question

Mastering Physics Solutions: Magnetic Field due to a Wire Conceptual Question

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

Part A = Bnet points out of the screen at A.
Part B = Bnet = 0 at B.
Part C = Bnet points into the screen at C.

The same amount of current I is flowing through two wires, labeled 1 and 2 in the figure, in the directions indicated by the arrows. In this problem you will determine the direction of the net magnetic field Bnet at each of the indicated points (A – C).
What is the direction of the magnetic field Bnet at point A? Recall that the currents in the two wires have equal magnitudes.
What is the direction of the magnetic field Bnet at point B?
What is the direction of the magnetic field Bnet at point C?

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Mastering Physics Solutions: ± Magnetic Force on a Current-Carrying Wire

Mastering Physics Solutions: ± Magnetic Force on a Current-Carrying Wire

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

Part A = 0.0000046 N
Part B = 0.083 N
Part C = due west
Part D = Current in the wire flows straight down; the magnetic field points due west.; Current in the wire flows straight up; the magnetic field points due east.; Current in the wire flows due east; the magnetic field points straight down.; Current in the wire flows due west and slightly up; the magnetic field points due east.
Part E = 0 N

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