At r = 0.766 R the magnetic field intensity will be half of its value at the center of the current carrying loop.
We have a circular loop of radius ' r ' carrying current ' i '.
We have to find at what distance along the axis of the loop is the magnetic field one-half its value at the center of the loop.
<h3>What is the formula to calculate the
Magnetic field intensity due to a current carrying circular loop at a point on its axis?</h3>
The formula to calculate the magnetic field intensity due to a current carrying ( i ) circular loop of radius ' R ' at a distance ' x ' on its axis is given by -

Now, for magnetic field intensity at the center of the loop can calculated by putting x = 0 in the above equation. On solving, we get -

Let us assume that the distance at which the magnetic field intensity is one-half its value at the center of the loop be ' r '. Then -




r = 0.766R
Hence, at r = 0.766 R - the magnetic field intensity will be half of its value at the center of the current carrying loop.
To solve more questions on magnetic field intensity, visit the link below-
brainly.com/question/15553675
#SPJ4
Answer:
28.7 m at 46.9°
Explanation:
The x component of the displacement is:
x = 6 m cos 0° + 25 m cos 57°
x = 19.6 m
The y component of the displacement is:
y = 6 m sin 0° + 25 m sin 57°
y = 21.0 m
The total displacement is found with Pythagorean theorem:
d = √(x² + y²)
d = 28.7 m
And the direction is found with trig:
θ = tan⁻¹(y/x)
θ = 46.9°
Answer:
I did this essay before, its not hard. Just read through it carefully and you'll figure it out. or just ask you mom, dad, guardian, or teacher for help
True is The answer would be I just did this