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MrRa [10]
3 years ago
14

The friends take a few minutes to discuss their ideas. Which of Amy's comments with respect to a charged particle moving in a ma

gnetic field is false? "The magnitude FB of the magnetic force exerted on the particle is proportional to the magnitude of the charge q and to the speed v of the particle." "The magnetic force acting on a charged particle moving in a magnetic field is perpendicular to the particle's velocity." "If the magnetic force is always perpendicular to the velocity, the path of the particle is a straight line." "The work done by the magnetic force on the particle is zero."
Physics
1 answer:
Marizza181 [45]3 years ago
7 0

Answer:

The statement "if the magnetic force is always perpendicular to the velocity, the path of the particle is a straight line" is false.

Explanation:

The equation for the magnetic force on a charge q moving at velocity v on a magnetic field B is given by the (vectorial) Lorentz Force Law F=qv\times B

From it we can clearly see that the <em>magnitude of the magnetic force </em>exerted on the particle is <em>proportional to the magnitude of the charge q and to the speed v of the particle</em>, and that it is also <em>perpendicular to the particle's velocity</em>. This means that at each instant it moves perpendicularly to the force, so <em>the work done by the magnetic force on the particle is zero</em>.

The statement "if the magnetic force is always perpendicular to the velocity, the path of the particle is a straight line" is false not only for this but for any force, a force always perpendicular to a velocity will curve the trajectory.

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Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

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Diameter for aluminum bar  = 40 mm

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The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

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i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

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Area of cross section with the hole is :

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A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

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