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3 years ago

Use the right-hand rule for magnetic force to determine the charge on the moving particle. This is a charge.

Physics

2 answers:

swat323 years ago

6 0

Answer:

The answer is negative

Explanation:

german3 years ago

4 0

Answer:

Use the right-hand rule for magnetic force to determine the charge on the moving particle.

This is a

negative

charge

Explanation:

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What is the magnitude of the force between two parallel wires 43.0 m long and 5.9 cm apart, each carrying 35.0 A in the same dir

beks73 [17]

Answer:

The magnitude of the force between the two wires is $17855.9*10^{-5} N$

Explanation:

If there are two parallel rectilinear conductors through which two electric currents of the same direction I1 and I2 circulate,both conductors will generate a magnetic field on each other, giving rise to a force between them.

To calculate the value of this force, first, according to the law of Biot and Savart, the magnetic field produced by conductor 1 over 2 is obtained, which will be given by the equation:

B1= $\frac{u_{o}*I1 }{2*\pi*a }$ equation 1

B1: Magnetic field produced by conductor 1

$u_{o}$ = free space permeability

a= distance between wires

I1= current carrying wire 1

This magnetic field exerts on a segment L of the conductor 2 through which a current of intensity I2 circulates, a force equal to:

F1-2= I2*L*B1 Equation2

We replaced B1 of the equation 1 in the equation 2:

F1-2= $I2*L*\frac{u_{o}*I1 }{2*\pi *a}$

F1-2= $\frac{u_{o}*I1*I2*L }{2*\pi *a}$

If we calculate the force exerted by conductor 2 on conductor 1 we would arrive at exactly the same value:

F2-1= F1-2

For this problem, the magnitude of the force between the two parallel cables that conduct current in the same direction is:

F1-2=F2-1=F

$u_{o} =4*\pi *10^{-7}$ Wb/A.m

I1=I1=35A

L=43M

a =5.9 cm= $5.9*10^{-2}$ m

$F=\frac{4*\pi *10^{-7}*35*35*43 }{2*\pi *5.9*10^{-2} }$

$F=17855.9*10^{-5} N$

Answer: The magnitude of the force between the two wires is $17855.9*10^{-5} N$

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A meter stick is pivoted at its 50 cm mark but does not balance because of non-uniformities in its material that cause its cente

Cerrena [4.2K]

Answer:

The mass of the stick is:

$M=0,9428kg$

The center of mass is at the 48,41cm mark

Explanation:

To be balanced in any case the resultant torque of the system must be zero.

For the first situation:

We have a 0,15kg mass al -0,4m from the pivot point, and a 0,3kg mass at +0,25m of the pivot point. On the other hand we know that the center of gravity of the stick is not at the 50cm mark point, so it will be at a X distance with a mass M, the torque sum:

$0,15kg*(-0,4m)+0,3kg*0,25m+X*M=0$

So:

(1) $X*M=-0,015kgm$

For the second situation:

Now the weights are interchanged, and as the pivot point has change, the distances from the pivot point are different:

We have a 0,3kg mass al -0,33m from the pivot point, and a 0,15kg mass at +0,32m of the pivot point. As the pivot has moved we have to move our reference for X: The center of gravity of the stick is at a X+0,07m distance with a mass M, the torque sum:

$0,3kg*(-0,33m)+0,15kg*0,32m+(X+0,07m)*M=0$