Illustrated · 1 min

Magnetic Field at Center of Loop

r is perpendicular to the direction of the current

Link to image above

dB=μo4π⋅Idl⃗sin⁡90r2dB = \frac{\mu_o}{4\pi} \cdot \frac{I \vec{dl} \sin{90}}{r^2} ∫dB=∫μoIdl⃗4πr2\int dB = \int \frac{\mu_oI\vec{dl}}{4\pi r^2} ∫dB=∫μoI4πr2∫dl⃗\int dB = \int \frac{\mu_oI}{4\pi r^2}\int \vec{dl} B=μoI4πR2(2πr)B = \frac{\mu_oI}{4\pi R^2}(2\pi r) B=μoI2rB = \frac{\mu_oI}{2r}

Magnetic Field on Loop Axis

Link to image above

r=x2+R2r = \sqrt{x^2+R^2}

sin⁡θ=Rr\sin \theta = \frac{R}{r} dB=μo4π⋅Idl⃗sin⁡θr2dB = \frac{\mu_o}{4\pi} \cdot \frac{I \vec{dl}\sin \theta}{r^2} dB=μo(I)4πr2⋅Rr⋅dldB = \frac{\mu_o(I)}{4\pi r^2}\cdot \frac{R}{r}\cdot dl dB=μo(I)(R)4πr3⋅dldB = \frac{\mu_o(I)(R)}{4\pi r^3}\cdot dl dB=μo(I)(R)4π(x2+R2)3/2⋅dldB = \frac{\mu_o(I)(R)}{4\pi (x^2 + R^2)^{3/2}} \cdot dl ∫dB=μo(I)(R)4π(x2+R2)3/2∫dl\int dB = \frac{\mu_o(I)(R)}{4\pi(x^2 + R^2)^{3/2}} \int dl B=μo(I)(R)4π(x2+R2)3/2(2πR)B = \frac{\mu_o(I)(R)}{4\pi(x^2 + R^2)^{3/2}} (2\pi R) B=μo(I)(R)22(x2+R2)3/2B = \frac{\mu_o(I)(R)^2}{2(x^2 + R^2)^{3/2}}

Example 1:

What is the magnetic field at the center of a thin, conducting ring with a diameter of 0.8m0.8m and a current of 2.2A2.2A?

B=μoI2rB = \frac{\mu_oI}{2r} B=(4π×10−7)(2.2)2(0.4)B = \frac{(4\pi \times 10^{-7})(2.2)}{2(0.4)} B≈3.456×10−6TB \approx 3.456 \times 10^{-6}T

Example 2:

A single-turn wire loop is 2.0cm2.0cm in diameter and carries a 650mA650mA current. Find the strength of the magnetic field (a) center of the loop and (b) on the loop axis, 20cm20cm from the center

(a)B=μoI2rB = \frac{\mu_oI}{2r} B=(4π×10−7)(0.65)2(0.01)B = \frac{(4\pi \times 10^{-7})(0.65)}{2(0.01)} B≈4.08×10−5TB \approx 4.08 \times 10^{-5}T

(b)B=μo(I)(R)22(x2+R2)3/2B = \frac{\mu_o(I)(R)^2}{2(x^2 + R^2)^{3/2}} B=(4π×10−7)(0.65)(0.01)22(0.202+0.012)3/2B = \frac{(4 \pi \times 10^{-7})(0.65)(0.01)^2}{2(0.20^2 + 0.01^2)^{3/2}} B≈5.086×10−9TB \approx 5.086 \times 10^{-9}T

Type a concept, like “slope” or “photosynthesis”.

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