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Anarel [89]
3 years ago
7

If the ball rolls out of bounds on the sideline in soccer, what is the result?

Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
6 0
The answer for this problem is C
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A Cassegrain telescope has a hole in the main mirror. This reduces the mirror area and thus the light gathering power, but how m
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Explanation: A cassegrain telescope is a kind of telescope which is made up of the curved mirrors one of the mirrors is a concave mirror is called the primary mirror and the second mirror called the secondary mirror which is a convex mirror, when light Penetrate the cassegrain telescope, it first hits the primary concave mirror and it's then reflected by the secondary convex mirror.

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Jan is holding an ice cube. What causes the ice to melt?
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A circular loop of radius r carries a current i. at what distance along the axis of the loop is the magnetic field one-half its
lana [24]

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 -

B(x) = \frac{\mu_{o} iR^{2} }{2(x^{2} +R^{2})^{\frac{3}{2} } }

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 -

B(0) = \frac{\mu_{o} i}{2R}

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 -

\frac{\mu_{o} iR^{2} }{2(r^{2} +R^{2})^{\frac{3}{2} } } = \frac{1}{2} \frac{\mu_{o}i }{2R}

2R^{3} = (r^{2} +R^{2} )^{\frac{3}{2} }

4R^{6} = (r^{2} +R^{2} )^{3}

r^{2} =0.587R^{2}

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

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