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Likurg_2 [28]
2 years ago
13

An electron traveling at a velocity v enters a uniform magnetic field B. Initially, the velocity and field are perpendicular to

each other. If the electron goes a distance d, the amount of work done on it by the magnetic field is ________.
Physics
1 answer:
Brut [27]2 years ago
6 0

If the electron goes a distance d, the amount of work done on it by the magnetic field is zero.

Because magnetic force acts perpendicular to the direction of motion, it has no effect on any moving charge particle. As a result, speed won't change.

<h3>What is Magnetic field?</h3>
  • The magnetic influence on moving electric charges, electric currents, and magnetic materials is described by a magnetic field, which is a vector field.
  • A force perpendicular to the charge's own velocity and the magnetic field acts on it when the charge is travelling through a magnetic field.
  • A compass, a motor, the magnets that hold items in refrigerators, railroad tracks, and modern roller coasters are examples of devices that use magnetic force.
  • A magnetic field is created by all moving charges, and any charges that move across its regions are subject to a force.

Learn more about Magnetic field here:

brainly.com/question/14848188

#SPJ4

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Answer:

W=17085KJ

Explanation:

From the question we are told that:

Height H=16m

Radius R=3

Height of water H_w=9m

Gravity g=9.8m/s

Density of water \rho=1000kg/m^3

Generally the equation for Volume of water is mathematically given by

 dv=\pi*r^2dy

 dv=\frac{\piR^2}{H^2}(H-y)^2dy

Where

   y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

 dw=(pdv)g (H-y)

Substituting dv

 dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)

 dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy

Therefore

 W=\int dw

 W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy

 W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)

 W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0

 W=3420.84*0.25[2401-65536]

 W=17084965.5J

 W=17085KJ

 

'

'

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