Magnetic Field at Center of Loop
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r = x 2 + R 2 r = \sqrt{x^2+R^2} r = x 2 + R 2
sin θ = R r \sin \theta = \frac{R}{r} sin θ = r R
d B = μ o 4 π ⋅ I d l ⃗ sin θ r 2 dB = \frac{\mu_o}{4\pi} \cdot \frac{I \vec{dl}\sin \theta}{r^2} d B = 4 π μ o ⋅ r 2 I d l sin θ
d B = μ o ( I ) 4 π r 2 ⋅ R r ⋅ d l dB = \frac{\mu_o(I)}{4\pi r^2}\cdot \frac{R}{r}\cdot dl d B = 4 π r 2 μ o ( I ) ⋅ r R ⋅ d l
d B = μ o ( I ) ( R ) 4 π r 3 ⋅ d l dB = \frac{\mu_o(I)(R)}{4\pi r^3}\cdot dl d B = 4 π r 3 μ o ( I ) ( R ) ⋅ d l
d B = μ o ( I ) ( R ) 4 π ( x 2 + R 2 ) 3 / 2 ⋅ d l dB = \frac{\mu_o(I)(R)}{4\pi (x^2 + R^2)^{3/2}} \cdot dl d B = 4 π ( x 2 + R 2 ) 3/2 μ o ( I ) ( R ) ⋅ d l
∫ d B = μ o ( I ) ( R ) 4 π ( x 2 + R 2 ) 3 / 2 ∫ d l \int dB = \frac{\mu_o(I)(R)}{4\pi(x^2 + R^2)^{3/2}} \int dl ∫ d B = 4 π ( x 2 + R 2 ) 3/2 μ o ( I ) ( R ) ∫ d l
B = μ o ( I ) ( R ) 4 π ( x 2 + R 2 ) 3 / 2 ( 2 π R ) B = \frac{\mu_o(I)(R)}{4\pi(x^2 + R^2)^{3/2}} (2\pi R) B = 4 π ( x 2 + R 2 ) 3/2 μ o ( I ) ( R ) ( 2 π R )
B = μ o ( I ) ( R ) 2 2 ( x 2 + R 2 ) 3 / 2 B = \frac{\mu_o(I)(R)^2}{2(x^2 + R^2)^{3/2}} B = 2 ( x 2 + R 2 ) 3/2 μ o ( I ) ( R ) 2
Example 1:
What is the magnetic field at the center of a thin, conducting ring with a diameter of 0.8 m 0.8m 0.8 m and a current of 2.2 A 2.2A 2.2 A ?
B = μ o I 2 r B = \frac{\mu_oI}{2r} B = 2 r μ o I
B = ( 4 π × 10 − 7 ) ( 2.2 ) 2 ( 0.4 ) B = \frac{(4\pi \times 10^{-7})(2.2)}{2(0.4)} B = 2 ( 0.4 ) ( 4 π × 1 0 − 7 ) ( 2.2 )
B ≈ 3.456 × 10 − 6 T B \approx 3.456 \times 10^{-6}T B ≈ 3.456 × 1 0 − 6 T
Example 2:
A single-turn wire loop is 2.0 c m 2.0cm 2.0 c m in diameter and carries a 650 m A 650mA 650 m A current. Find the strength of the magnetic field (a) center of the loop and (b) on the loop axis, 20 c m 20cm 20 c m from the center
(a)B = μ o I 2 r B = \frac{\mu_oI}{2r} B = 2 r μ o I
B = ( 4 π × 10 − 7 ) ( 0.65 ) 2 ( 0.01 ) B = \frac{(4\pi \times 10^{-7})(0.65)}{2(0.01)} B = 2 ( 0.01 ) ( 4 π × 1 0 − 7 ) ( 0.65 )
B ≈ 4.08 × 10 − 5 T B \approx 4.08 \times 10^{-5}T B ≈ 4.08 × 1 0 − 5 T
(b)B = μ o ( I ) ( R ) 2 2 ( x 2 + R 2 ) 3 / 2 B = \frac{\mu_o(I)(R)^2}{2(x^2 + R^2)^{3/2}} B = 2 ( x 2 + R 2 ) 3/2 μ o ( I ) ( R ) 2
B = ( 4 π × 10 − 7 ) ( 0.65 ) ( 0.01 ) 2 2 ( 0.20 2 + 0.01 2 ) 3 / 2 B = \frac{(4 \pi \times 10^{-7})(0.65)(0.01)^2}{2(0.20^2 + 0.01^2)^{3/2}} B = 2 ( 0.2 0 2 + 0.0 1 2 ) 3/2 ( 4 π × 1 0 − 7 ) ( 0.65 ) ( 0.01 ) 2
B ≈ 5.086 × 10 − 9 T B \approx 5.086 \times 10^{-9}T B ≈ 5.086 × 1 0 − 9 T