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Rom4ik [11]
3 years ago
13

Consider an electron traveling horizontally in the positive direction, above, near, and parallel to a current-carrying wire. The

current is traveling in the negative direction.
(a) Will the electron keep on going straight, move towards the wire or away from the wire?


(b) If we replace the electron with a proton and now the proton is coming from above, traveling in the perpendicular direction, moving towards the wire. Will the proton hit the wire, veer to the left? veer to the right? stop? go out of the page? go onto the page?
Physics
1 answer:
allsm [11]3 years ago
6 0

Answer:

(a) The electron will move towards the wire.

The direction of the magnetic fields created by the wire can be found via right-hand rule. If you point your thumb towards the direction of the current, and if you curl your fingers, the direction of your four fingers will give the direction of the magnetic field. In this case, magnetic field is around the wire, and into the page just above the wire, where the electron is located.

\vec{F} = q\vec{v} \times \vec{B}

According to the above formula, the direction of the force the wire applies to the electron can be found by right-hand rule.

Since the electron has a negative charge, the direction of the force is towards the wire.

(b) The proton will veer to the right.

The direction of the magnetic field is the same as the previous part. The proton has a positive charge, and coming from above. The direction of its velocity is downwards. The magnetic field above the wire is pointed into the page. Using the right-hand rule, the magnetic force on the proton is directed to the right, with respect to us.

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Let  us consider two bodies having masses m and m' respectively.

Let they are  separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as -  F = G\frac{mm'}{r^{2} }   where G is the gravitational force constant.

From the above we see that F ∝ mm' and F\alpha \frac{1}{r^{2} }

Let the orbital radius of planet  A is r_{1}  = r and mass of planet is m_{1}.

Let the mass of central star is m .

Hence the gravitational force for planet A  is f_{1} =G \frac{m_{1}*m }{r^{2} }

For planet B the orbital radius  r_{2} =2r_{1} and mass m_{2} = 3 m_{1}

Hence the gravitational force f_{2} =G\frac{m m_{2} }{r^{2} }

                                                 f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }

                                                 = \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }

Hence the ratio is  \frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2}  }{Gmm_{1}/r_{1} ^2 }

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3 years ago
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Explanation:

It is given that,

Mass of the woman, m₁ = 52 kg

Angular velocity, \omega=0.47\ rev/s=2.95\ rad/s

Mass of disk, m₂ = 118 kg

Radius of the disk, r = 3.9 m

The moment of inertia of woman which is standing at the rim of a large disk is :

I={m_1r^2}

I={52\times 3.9^2}

I₁ = 790.92 kg-m²

The moment of inertia of of the disk about an axis through its center is given by :

I_2=\dfrac{m_2r^2}{2}

I_2=\dfrac{118\times (3.9)^2}{2}

I₂ =897.39 kg-m²

Total moment of inertia of the system is given by :

I=I_1+I_2

I=790.92+897.39

I = 1688.31 kg-m²

The angular momentum of the system is :

L=I\times \omega

L=1688.31 \times 2.95

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So, the total angular momentum of the system is 4980.5 kg-m²/s. Hence, this is the required solution.

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Answer

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Hence;

F = 0.5 x 100 x 9.8

F = 490N.

We will only need the coefficient of kinetic friction if we were asked to find the force required to keep the object moving uniformly. Usually, the force needed to keep an object moving uniformly over a rough surface is lesser that which is needed to start its motion.

In this problem, we were only asked to find the minimum force required to make the object move which we have done.

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