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Slav-nsk [51]
3 years ago
6

The two long straight wires are separated by a distance of d = 0.40 m. The currents are to the right in the upper wire and to th

e left in the lower wire. What is the direction of the magnetic field at point P, that is a distance d/2 below the lower wire?

Physics
1 answer:
elena-s [515]3 years ago
4 0

Answer:

Out of the page

Explanation:

At point P, Using right hand rule, the direction of magnetic field  due to current flowing towards left in lower wire comes out to be in outward direction.

At point P, Using right hand rule, the direction of magnetic field  due to current flowing towards right in upper wire comes out to be in inward direction.

we know that magnetic field at a point by a long current carrying wire depends inversely on the distance from the wire. hence greater the distance, weaker will be the magnetic field.

Since point P is farther from upper wire and closer to lower wire, the magnetic field in Outward direction by the lower wire is stronger as compared to the magnetic field by the upper wire in inward direction.

hence magnetic field at point P comes out to be in Outward direction.

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high school physics, no need detail explain, just give the answer, but you have to make sure thank you
andrey2020 [161]

Answer:

approximately 30 degrees

Explanation:

If it takes the cannonball 2 seconds to reach the maximum height, we can use the analysis of the vertical component of the velocity and the fact that the acceleration of gravity is the one acting opposite to this initial vertical component (v_y) of the velocity. We know as well that at the top of the trajectory, the vertical component of the velocity is zero, and then the movement starts going down in it trajectory. So, the final velocity for the first part of the ascending movement is zero, giving us the following equation for the velocity under an accelerated movement (with acceleration of gravity "g" acting):

v_f=v_i-a\,t\\v_f=v_y-g\,t\\0=v_y-9.8\,*\,2\\v_y=9.8\,*\,2=19.6 \frac{m}{s}

By knowing the vertical component of the initial velocity (19.6 m/s), and the actual magnitude of the total initial velocity (40 m/s), we can calculate what angle was the initial velocity vector forming above the horizontal. We use for such the fact that the sine of the angle relates the opposite side of a right angle triangle with the hypotenuse, and solve for the angle using the arcsin function:

sin(\theta)=\frac{opp}{hyp} \\sin(\theta)=\frac{19.6}{40}\\\theta=arcsin(\frac{19.6}{40})\\\theta=29.34^o

which tells us that the closer answer shown is 30^o

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3 years ago
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Answer:

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2 years ago
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