Answer :
The magnetic field at the center of the 2nd loop is half the magnetic field at the center of the 1st loop.
Explanation:
The magnetic field strength at the center of a current-carrying circular loop is
[tex]B=\frac{\mu_0 I}{2R}[/tex]
where
[tex]\mu_0[/tex] is the vacuum permeability
I is the current
R is the radius of the loop
For the first circular loop, we have
[tex]B_1 = \frac{\mu_0 I_1}{2R_1}[/tex]
The second loop has
[tex]I_2 = I_1[/tex] (same current)
[tex]R_2 = 2R_1[/tex] (twice the radius)
So, the magnetic field at the centre of the second loop is
[tex]B_2 = \frac{\mu_0 I_2}{2R_2}=\frac{1}{2}(\frac{\mu_0 I_1}{2R_1})=\frac{B_1}{2}[/tex]
So, the magnetic field at the center of the 2nd loop is half the magnetic field at the center of the 1st loop.
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The magnetic fields at the centers of the loops two is half compare to the magnetic filed of at the center of loop 1.
What is magnetic field?
The magnetic field is the field in the space and around the magnet in which the magnetic field can be fill.
The magnetic filed at center of the loop can be given as,
[tex]B=\dfrac{\mu_o I}{2R}[/tex]
Here, ([tex]\mu_o[/tex]) is the magnetic constant, (R) radius of the loop and (I) is the current carried by it.
As the circular loop one has the radius (say R). Thus the magnetic field for circular loop one can be given as,
[tex]B_1=\dfrac{\mu_o I}{2R}[/tex]
Now, the radius of loop two is twice that of the radius of loop one. Thus the magnetic field for circular loop two can be given as,
[tex]B_2=\dfrac{\mu_o I}{2\times2R}\\B_2=\dfrac{1}{2}\times\dfrac{\mu_o I}{2R}\\B_2=\dfrac{1}{2}\times B_1[/tex]
Thus the magnetic fields at the centers of the loops two is half compare to the magnetic filed of at the center of loop 1.
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