Four long, parallel conductors carry equal currents of I = 5.00 A. The direction of the current is into the page at points A and
B (indicated by the crosses) and out of the page at C and D (indicated by the dots). Calculate the magnitude and direction of the magnetic field at point P, located at the center of the square with an edge length 0.200 m.
1 answer:
Answer:
Explanation:
The magnetic field due to a straight wire carrying current is given by
B = μo• I / 2πr
Where,
μo = permeability of free space
μo = 4π×10^-7 Tm/A
I is the current in wire
I = 5A
r Is the distance of point from the wire.
The distance between the parallel conductors and the point is r/2
R = 0.283/2 = 0.1415m
Check attachment on how I used Pythagoras theorem to find the diagonal of the square..
Hence, the magnetic field at point P is
B = μo•I / 2πR
B = 4π × 10^-7 × 5 / 2π × 0.1415
B = 7.07 × 10^-6 T.
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Answer:
Explanation:
This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.
The initial velocity is in the x-direction, and there is no acceleration in the x-direction.
On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.
Applying the equations of kinematics in the x-direction gives
For the y-direction gives
Combining both equation yields the y_component of the final velocity
Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.