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11. Two concentric loops (each with only 1 turn) carry the currents shown below. What is the magnitude of the magnetic field at the center of both loops. R 2I I 2R a) \frac{\mu_0 I}{2R} b) \frac{5\mu_0 I}{4R} c) \frac{3\mu_0 I}{4R} d) \frac{2\mu_0 I}{R} e) None of the answers is correct

          11. Two concentric loops (each with only 1 turn) carry the currents shown below. What is the
magnitude of the magnetic field at the center of both loops.
R
2I
I
2R
a) \frac{\mu_0 I}{2R}
b) \frac{5\mu_0 I}{4R}
c) \frac{3\mu_0 I}{4R}
d) \frac{2\mu_0 I}{R}
e) None of the answers is correct
        
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11. Two concentric loops (each with only 1 turn) carry the currents shown below. What is the
magnitude of the magnetic field at the center of both loops.
R
2I
I
2R
a) (μ0 I)/(2R)
b) (5μ0 I)/(4R)
c) (3μ0 I)/(4R)
d) (2μ0 I)/(R)
e) None of the answers is correct

Added by Jamie N.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Two concentric loops (each with only 1 turn) carry the currents shown below. What is the magnitude of the magnetic field at the center of both loops? R 2√1 2R oI a) 2R 501 b) 4R 301 c) 4R 201 d) R e) None of the answers is correct
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Transcript

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00:01 Hello students, in this question we have given two currents.
00:06 So let's take i1 is given as 5i and i2 is given as 19i.
00:13 Now and we need to find the ratio of the magnetic field created by loop 1 of v1 and loop 2 for v2 at the center of this first loop that is from point c.
00:28 So basically we need to find v1 upon v2 as a ratio.
00:34 Now to find ratio first the magnetic field point c due to loop 1.
00:50 So here v1 magnetic field will be equal to mu knot i upon 2r and that is towards left.
01:03 Now for magnetic field, magnetic field at point c due to loop 2 that's v2 magnetic field v2.
01:23 So we can write expression for this v2 is equal to mu knot i2 r2 r square divided by 2r square plus r square whole power 3 by 2 that is towards right.
01:46 Now this is the magnetic field on the axis of loop, axis of loop.
01:55 Now this can be written as further v2 will be equal to mu knot i2 r square divided by 2 into 2r square power 3 by 2.
02:10 Now for the solving this we can write magnetic field v2 so this will be equal to mu knot i2 r square upon 4 root 2 r cube...
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