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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We don't know that at all. The 3rd law says that the REaction is opposite and EQUAL to the action. We don't know where that "twice as much" comes from.
The tension in the string and the acceleration must be equal for both masses. (See the free body diagrams)
Answer:
<h2>4.9 J</h2>
Explanation:
The gravitational potential energy of a body can be found by using the formula
GPE = mgh
where
m is the mass
h is the height
g is the acceleration due to gravity which is 9.8 m/s²
From the question we have
GPE = 10 × 9.8 × 0.05
We have the final answer as
<h3>4.9 J</h3>
Hope this helps you