A small disk of radius R1 is mounted coaxially with a larger disk of radius R2. The disks are securely fastened to each other an
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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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To solve this problem we will consider the concepts related to the normal deformation on a surface, generated when the change in length is taken per unit of established length, that is, the division between the longitudinal fraction gained or lost, over the initial length. In general mode this normal deformation can be defined as
Here,
= Change in final length
and the initial length
PART A)
PART B)
PART C)
Therefore the rank of this deformation would be B>C>A