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.
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Answer:
<em>displacement = -85 miles</em>
Explanation:
<u>Displacement
</u>
It's a magnitude used to measure the linear space between two points. It's computed as the subtraction of the final position minus the initial position which results in a vector. Notice the displacement only depends on the initial and final positions and not on the path the object has traveled.
Brayden starts to measure his position when the mile marker reads 260. Then he travels to the 150-mile marker and goes back to the 175-mile marker, his final position. As mentioned, the displacement only depends on the relative positions, so
displacement = 175 - 260 = -85 miles
Answer:
P /K = 1,997 10⁻³⁶ s⁻¹
Explanation:
For this exercise let's start by finding the radiation emitted from the accelerator
= 
the radius of the orbit is the radius of the accelerator a = r = 0.530 m
let's calculate
\frac{dE}{dt} = [(1.6 10⁻¹⁹)² 0.530²] / [6π 8.85 10⁻¹² (3 108)³]
P= \frac{dE}{dt}= 1.597 10⁻⁵⁴ W
Now let's reduce the kinetic energy to SI units
K = 5.0 10⁶ eV (1.6 10⁻¹⁹ J / 1 eV) = 8.0 10⁻¹⁹ J
the fraction of energy emitted is
P / K = 1.597 10⁻⁵⁴ / 8.0 10⁻¹⁹
P /K = 1,997 10⁻³⁶ s⁻¹
Answer:
M_c = 100.8 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two cases shown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *144
M_c = 2.5*( 42 / 150 ) *144
M_c = 100.8 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
Answer:
Explanation:
(a) The velocity of object is zero when it is at maximum height.
(b) The direction of velocity changes as it starts moving downwards after it reaches the maximum height.
(c) Acceleration due to gravity always acts downwards so its sign remains same.
