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valentina_108 [34]
3 years ago
9

A conductor carrying a current I = 16.5 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A

magnetic force per unit length of 0.130 N/m acts on the conductor in the negative y direction. (a) Determine the magnitude of the magnetic field in the region through which the current passes.
Physics
1 answer:
Jet001 [13]3 years ago
6 0

To solve this problem we will apply the concepts related to the Magnetic Force, this is given by the product between the current, the body length, the magnetic field and the angle between the force and the magnetic field, mathematically that is,

F = ILBsin \theta

Here,

I = Current

L = Length

B = Magnetic Field

\theta = Angle between Force and Magnetic Field

But \theta = 90\°

F = ILB

Rearranging to find the Magnetic Field,

B = \frac{F}{IL}

Here the force per unit length,

B = \frac{1}{I}\frac{F}{L}

Replacing with our values,

B = \frac{0.130N/m}{16.5}

B = 0.0078T

Therefore the magnitude of the magnetic field in the region through which the current passes is 0.0078T

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Determine the components of reaction at the fixed support
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The components of reaction at the fixed support are  

    A_{(x)}  = 400  \ N ,  A_{(y)}  = -500  \ N ,  A_{(z)}  = 600  \ N ,  M_x  = 1225 \  N\cdot m , M_y  = 750 \  N\cdot m ,  M_z  = 0 \  N\cdot m

Explanation:

Looking at  the diagram uploaded we see that there are two  forces acting along the x-axis on the fixed support    

   These force are  400 N  and  A_{(x)} [ i.e the reactive force of  400 N  ]

Hence the sum of forces along the x axis is mathematically represented as

        A_{(x)}  - 400  = 0

=>     A_{(x)}  = 400  \ N

Looking at  the diagram uploaded we see that there are two  forces acting along the y-axis on the fixed support  

   These force are  500 N  and  A_{(y)} [ i.e the force acting along the same direction with 500 N   ]

Hence the sum of forces along the x axis is mathematically represented as

        A_{(y)}  + 500  = 0

=>     A_{(y)}  = -500  \ N

Looking at  the diagram uploaded we see that there are two  forces acting along the z-axis on the fixed support  

       These force are  600 N  and  A_{(z)} [ i.e the reactive force of  600 N  ]

Hence the sum of forces along the x axis is mathematically represented as

        A_{(z)}  - 600  = 0

=>     A_{(z)}  = 600  \ N

Generally taking moment about A along the x-axis we have that

    \sum M_x  = M_x  - 500 (0.75 + 0.5) + 600 ( 1 ) = 0

=>   M_x  = 1225 \  N\cdot m

Generally taking moment about A along the y-axis we have that

    \sum M_y  = M_y  - 400 (0.75 ) + 600 ( 0.75 ) = 0

=>   M_y  = 750 \  N\cdot m

Generally taking moment about A along the z-axis we have that

    \sum M_z  = M_z = 0

=>   M_z  = 0 \  N\cdot m

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Answer:

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Explanation:

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