Magnetostática Estudia los campos magnéticos generados por corrientes estacionarias. En este régimen no hay variación temporal del campo eléctrico, por lo que las únicas fuentes del campo magnético son las corrientes eléctricas.
Ecuaciones de Maxwell Ley de Gauss Establece que la divergencia del campo magnético es cero, lo que implica que no existen monopolos magnéticos.
∇ ⋅ B ⃗ = 0 \Large \nabla \cdot \vec{B} = 0 ∇ ⋅ B = 0 ∮ S B ⃗ ⋅ d S = 0 \Large \oint_{S} \vec{B}\cdot d\mathbf{S} = 0 ∮ S B ⋅ d S = 0 Ley de Ampère Relaciona el rotor del campo magnético con la densidad de corriente estacionaria, indicando que las corrientes producen líneas de campo cerrado.
∇ × B ⃗ = μ J ⃗ \Large \nabla \times \vec{B} = \mu\,\vec{J} ∇ × B = μ J ∮ C B ⃗ ⋅ d ℓ = μ I \Large \oint_{C} \vec{B}\cdot d\boldsymbol{\ell} = \mu\,I ∮ C B ⋅ d ℓ = μ I Función de Green y su operador diferencial Sea L ( r ) L(\mathbf{r}) L ( r ) G ( r , r ′ ) G(\mathbf{r},\mathbf{r'}) G ( r , r ′ )
L ( r ⃗ ) G ( r ⃗ − r ′ ⃗ ) = δ ( 3 ) ( r ⃗ − r ′ ⃗ ) \Large L(\vec{r})\,\textcolor{#ae58ff}{G(\vec{r}-\vec{r'})} = \delta^{(3)}(\vec{r} - \vec{r'}) L ( r ) G ( r − r ′ ) = δ ( 3 ) ( r − r ′ ) Multiplicando por una función x ( r ′ ) x(\mathbf{r'}) x ( r ′ )
L ( r ⃗ ) [ G ( r ⃗ − r ′ ⃗ ) x ( r ′ ⃗ ) ] = δ ( 3 ) ( r ⃗ − r ′ ⃗ ) x ( r ′ ⃗ ) \Large L(\vec{r})\,\bigl[\textcolor{#ae58ff}{G(\vec{r}-\vec{r'})}\,x(\vec{r'})\bigr] = \delta^{(3)}(\vec{r} - \vec{r'})\,x(\vec{r'}) L ( r ) [ G ( r − r ′ ) x ( r ′ ) ] = δ ( 3 ) ( r − r ′ ) x ( r ′ ) Integrando sobre el dominio Ω \Omega Ω
∫ Ω L ( r ⃗ ) [ G ( r ⃗ − r ′ ⃗ ) x ( r ′ ⃗ ) ] d Ω = ∫ Ω δ ( 3 ) ( r ⃗ − r ′ ⃗ ) x ( r ′ ⃗ ) d Ω \Large \int_{\Omega} L(\vec{r})\,\bigl[\textcolor{#ae58ff}{G(\vec{r}-\vec{r'})}\,x(\vec{r'})\bigr]\,d\Omega = \int_{\Omega} \delta^{(3)}(\vec{r} - \vec{r'})\,x(\vec{r'})\,d\Omega ∫ Ω L ( r ) [ G ( r − r ′ ) x ( r ′ ) ] d Ω = ∫ Ω δ ( 3 ) ( r − r ′ ) x ( r ′ ) d Ω Por linealidad, esto da directamente la solución de L ϕ = x L\phi = x L ϕ = x
L ( r ⃗ ) [ ∫ Ω G ( r ⃗ − r ′ ⃗ ) x ( r ′ ⃗ ) d Ω ] = x ( r ⃗ ) \Large L(\vec{r}) \left[ \int_{\Omega} \textcolor{#ae58ff}{G(\vec{r}-\vec{r'})}\,x(\vec{r'})\,d\Omega \right] = x(\vec{r}) L ( r ) [ ∫ Ω G ( r − r ′ ) x ( r ′ ) d Ω ] = x ( r ) El potencial Vectorial Magnético Como ∇ ⋅ B ⃗ = 0 \nabla \cdot \vec{B} = 0 ∇ ⋅ B = 0 A ⃗ \vec{A} A
B ⃗ = ∇ × A ⃗ \vec{B} = \nabla \times \vec{A} B = ∇ × A De la ley de Ampère en forma diferencial para campo magnético:
∇ × B ⃗ = μ J ⃗ \nabla \times \vec{B} = \mu \,\vec{J} ∇ × B = μ J ∇ × ( ∇ × A ⃗ ) = μ J ⃗ \nabla \times (\nabla \times \vec{A}) = \mu \,\vec{J} ∇ × ( ∇ × A ) = μ J ∇ 2 v ⃗ = ∇ ( ∇ ⋅ v ⃗ ) − ∇ × ( ∇ × v ⃗ ) \large \nabla^2 \,\vec{v} = \nabla \bigl(\nabla \cdot \vec{v}\bigr) \;-\; \nabla \times \bigl(\nabla \times \vec{v}\bigr) ∇ 2 v = ∇ ( ∇ ⋅ v ) − ∇ × ( ∇ × v ) Aplicando la identidad vectorial sobre A ⃗ \vec{A} A
∇ 2 A ⃗ = ∇ ( ∇ ⋅ A ⃗ ) − ∇ × ( ∇ × A ⃗ ) \nabla^2 \,\vec{A} = \nabla ( \nabla \cdot \vec{A} ) - \nabla \times ( \nabla \times \vec{A} ) ∇ 2 A = ∇ ( ∇ ⋅ A ) − ∇ × ( ∇ × A ) ∇ 2 A ⃗ = ∇ ( ∇ ⋅ A ⃗ ) − ∇ × ( B ⃗ ) \nabla^2 \,\vec{A} = \nabla ( \nabla \cdot \vec{A} ) - \nabla \times ( \vec{B} ) ∇ 2 A = ∇ ( ∇ ⋅ A ) − ∇ × ( B ) ∇ 2 A ⃗ = ∇ ( ∇ ⋅ A ⃗ ) − μ J ⃗ \nabla^2 \,\vec{A} = \nabla ( \nabla \cdot \vec{A} ) - \mu \,\vec{J} ∇ 2 A = ∇ ( ∇ ⋅ A ) − μ J Gauge de Coulomb
∇ ⋅ A ⃗ = 0 \large \nabla \cdot \vec{A} = 0 ∇ ⋅ A = 0 Con esto, la identidad anterior se simplifica:
∇ 2 A ⃗ = ∇ ( ∇ ⋅ A ⃗ ) − μ J ⃗ \nabla^2 \,\vec{A} = \cancel{\nabla ( \nabla \cdot \vec{A} )} - \mu \,\vec{J} ∇ 2 A = ∇ ( ∇ ⋅ A ) − μ J ∇ 2 A ⃗ = − μ J ⃗ \nabla^2 \,\vec{A} = - \mu \,\vec{J} ∇ 2 A = − μ J Resolviendo la ecuación de Poisson vectorial ∇ 2 A ⃗ = − μ J ⃗ \Huge \nabla^2 \,\vec{A} = - \mu \,\vec{J} ∇ 2 A = − μ J L ( r ⃗ ) y ( r ⃗ ) = x ( r ⃗ ) \Large L(\vec{r}) y(\vec{r}) = x(\vec{r}) L ( r ) y ( r ) = x ( r ) L ( r ⃗ ) G ( r ⃗ − r ′ ⃗ ) = δ ( 3 ) ( r ⃗ − r ′ ⃗ ) \Large L(\vec{r})\,\textcolor{#ae58ff}{G(\vec{r}-\vec{r'})} = \delta^{(3)}(\vec{r} - \vec{r'}) L ( r ) G ( r − r ′ ) = δ ( 3 ) ( r − r ′ ) y ( r ⃗ ) = [ ∫ Ω G ( r ⃗ − r ′ ⃗ ) x ( r ′ ⃗ ) d Ω ] \Large y(\vec{r})= \left[ \int_{\Omega}\textcolor{#ae58ff}{G(\vec{r}-\vec{r'})}\,x(\vec{r'}) d\Omega \right]\, y ( r ) = [ ∫ Ω G ( r − r ′ ) x ( r ′ ) d Ω ] ∇ 2 A ⃗ = − μ J ⃗ \Large \nabla^2 \,\vec{A} = - \mu \,\vec{J} ∇ 2 A = − μ J ∇ 2 [ − 1 4 π 1 ∣ r ⃗ − r ′ ⃗ ∣ ] = δ ( 3 ) ( r ⃗ − r ′ ⃗ ) \Large \nabla^2 \textcolor{#ae58ff}{\left[ -\frac{1}{4\pi} \frac{1}{| \vec{r}-\vec{r'} |} \right]} = \delta^{(3)}(\vec{r} - \vec{r'}) ∇ 2 − 4 π 1 ∣ r − r ′ ∣ 1 = δ ( 3 ) ( r − r ′ ) A ( r ⃗ ) = [ ∫ Ω [ − 1 4 π 1 ∣ r ⃗ − r ′ ⃗ ∣ ] ( − μ J ⃗ ) d Ω ] \Large A(\vec{r})= \left[ \int_{\Omega} \textcolor{#ae58ff}{\left[ -\frac{1}{4\pi} \frac{1}{| \vec{r}-\vec{r'} |} \right]} \, (- \mu \,\vec{J}) d\Omega \right]\, A ( r ) = ∫ Ω − 4 π 1 ∣ r − r ′ ∣ 1 ( − μ J ) d Ω A ( r ⃗ ) = μ 4 π ∫ Ω J ⃗ ( r ′ ⃗ ) ∣ r ⃗ − r ′ ⃗ ∣ d Ω \Large A(\vec{r})= \frac{\mu }{4\pi} \int_{\Omega} \frac{\vec{J}(\vec{r'})}{| \vec{r}-\vec{r'} |} \, d\Omega \, A ( r ) = 4 π μ ∫ Ω ∣ r − r ′ ∣ J ( r ′ ) d Ω Campo Magnético resultante B ⃗ = ∇ × A ⃗ \Huge \vec{B} = \nabla \times \vec{A} B = ∇ × A B ⃗ = ∇ × μ 4 π ∫ Ω J ⃗ ( r ′ ⃗ ) ∣ r ⃗ − r ′ ⃗ ∣ d Ω \Large \vec{B} = \nabla \times \frac{\mu }{4\pi} \int_{\Omega} \frac{\vec{J}(\vec{r'})}{| \vec{r}-\vec{r'} |} \, d\Omega \, B = ∇ × 4 π μ ∫ Ω ∣ r − r ′ ∣ J ( r ′ ) d Ω B ⃗ = μ 4 π ∫ Ω ∇ × J ⃗ ( r ′ ⃗ ) ∣ r ⃗ − r ′ ⃗ ∣ d V ′ \Large \vec{B} = \frac{\mu }{4\pi} \int_{\Omega} \nabla \times \frac{\vec{J}(\vec{r'})}{| \vec{r}-\vec{r'} |} \, d V' \, B = 4 π μ ∫ Ω ∇ × ∣ r − r ′ ∣ J ( r ′ ) d V ′ ∇ × c ⃗ ∣ r ⃗ − r ′ ⃗ ∣ = c ⃗ × r ⃗ − r ′ ⃗ ∣ r ⃗ − r ′ ⃗ ∣ 3 \large \nabla \times \frac{\vec{c}}{| \vec{r}-\vec{r'} |} = \vec{c} \times \frac{\vec{r}-\vec{r'}}{| \vec{r}-\vec{r'} |^{3}} ∇ × ∣ r − r ′ ∣ c = c × ∣ r − r ′ ∣ 3 r − r ′ B ⃗ = μ 4 π ∫ Ω J ⃗ ( r ′ ⃗ ) × r ⃗ − r ′ ⃗ ∣ r ⃗ − r ′ ⃗ ∣ 3 d V ′ \Large \vec{B} = \frac{\mu }{4\pi} \int_{\Omega} \vec{J}(\vec{r'}) \times \frac{\vec{r}-\vec{r'}}{| \vec{r}-\vec{r'} |^{3}} \, d V' \, B = 4 π μ ∫ Ω J ( r ′ ) × ∣ r − r ′ ∣ 3 r − r ′ d V ′ Potencial vectorial A ⃗ ( r ⃗ ) = μ 4 π ∫ Ω J ⃗ ( r ′ ⃗ ) ∣ r ⃗ − r ′ ⃗ ∣ d Ω \Large \vec{A}(\vec{r})= \frac{\mu }{4\pi} \int_{\Omega} \frac{\vec{J}(\vec{r'})}{| \vec{r}-\vec{r'} |} \, d\Omega \, A ( r ) = 4 π μ ∫ Ω ∣ r − r ′ ∣ J ( r ′ ) d Ω B ⃗ = ∇ × A ⃗ \large \vec{B} = \nabla \times \vec{A} B = ∇ × A Fórmula para corrientes en hilo A ⃗ ( r ⃗ ) = μ 4 π ∮ L I d ℓ ′ ⃗ ∣ r ⃗ − r ′ ⃗ ∣ \Large \vec{A}(\vec{r}) = \frac{\mu }{4\pi} \oint_{L} \frac{I d\vec{ℓ'} }{| \vec{r}-\vec{r'} |} A ( r ) = 4 π μ ∮ L ∣ r − r ′ ∣ I d ℓ ′ Campo directo B ⃗ ( r ⃗ ) = μ 4 π ∫ Ω J ⃗ ( r ′ ⃗ ) × [ r ⃗ − r ′ ⃗ ] ∣ r ⃗ − r ′ ⃗ ∣ 3 d V ′ \Large \vec{B}(\vec{r}) = \frac{\mu }{4\pi} \int_{\Omega} \frac{\vec{J}(\vec{r'}) \times [\vec{r}-\vec{r'}]}{| \vec{r}-\vec{r'} |^{3}} \, d V' \, B ( r ) = 4 π μ ∫ Ω ∣ r − r ′ ∣ 3 J ( r ′ ) × [ r − r ′ ] d V ′ Ley de Biot Savart Fórmula para corrientes en hilo B ⃗ ( r ⃗ ) = μ 4 π ∮ L I d ℓ ′ ⃗ × [ r ⃗ − r ′ ⃗ ] ∣ r ⃗ − r ′ ⃗ ∣ 3 \Large \vec{B}(\vec{r}) = \frac{\mu }{4\pi} \oint_{L} \frac{I d\vec{ℓ'} \times[\vec{r}-\vec{r'}]}{| \vec{r}-\vec{r'} |^{3}} B ( r ) = 4 π μ ∮ L ∣ r − r ′ ∣ 3 I d ℓ ′ × [ r − r ′ ] Ejemplos Hilo recto infinito I \Huge I I Potencial vectorial A ⃗ ( r ⃗ ) = μ 0 4 π ∮ L I d ℓ ⃗ ′ ∣ r ⃗ − r ⃗ ′ ∣ \displaystyle \vec{A}(\vec{r}) \;=\; \frac{\mu_0}{4\pi} \oint_{L} \frac{I\,d\vec{\ell}'}{\lvert \vec{r} - \vec{r}'\rvert} A ( r ) = 4 π μ 0 ∮ L ∣ r − r ′ ∣ I d ℓ ′ Para r ⃗ = ( r , 0 , 0 ) , r ⃗ ′ = ( 0 , 0 , z ′ ) , d ℓ ⃗ ′ = z ^ d z ′ \text{Para } \vec{r} = (r,0,0),\; \vec{r}'=(0,0,z'),\; d\vec{\ell}'=\hat{z}\,dz' Para r = ( r , 0 , 0 ) , r ′ = ( 0 , 0 , z ′ ) , d ℓ ′ = z ^ d z ′ A ⃗ ( r ) = z ^ μ 0 I 4 π ∫ − ∞ + ∞ d z ′ r 2 + ( z ′ ) 2 = z ^ μ 0 I 2 π ln ( r ) \displaystyle \vec{A}(r)
= \hat{z}\,\frac{\mu_0\,I}{4\pi} \int_{-\infty}^{+\infty} \frac{dz'}{\sqrt{r^2 + (z')^2}}
= \hat{z}\,\frac{\mu_0\,I}{2\pi} \ln(r) \; A ( r ) = z ^ 4 π μ 0 I ∫ − ∞ + ∞ r 2 + ( z ′ ) 2 d z ′ = z ^ 2 π μ 0 I ln ( r ) B ⃗ = ∇ × A ⃗ \Large \vec{B} = \nabla \times \vec{A} B = ∇ × A B φ ( r ) = − d d r ( μ 0 I 2 π ln ( r ) ) = μ 0 I 2 π r \displaystyle B_{\varphi}(r)
= -\,\frac{d}{dr}\Bigl(\tfrac{\mu_0\,I}{2\pi}\ln(r)\Bigr)
= \frac{\mu_0\,I}{2\pi\,r} B φ ( r ) = − d r d ( 2 π μ 0 I ln ( r ) ) = 2 π r μ 0 I Campo directo (Biot–Savart) B ⃗ ( r ⃗ ) = μ 0 4 π ∮ L I d ℓ ⃗ ′ × ( r ⃗ − r ⃗ ′ ) ∣ r ⃗ − r ⃗ ′ ∣ 3 \displaystyle \vec{B}(\vec{r})
= \frac{\mu_0}{4\pi} \oint_{L} \frac{I\,d\vec{\ell}' \times (\vec{r} - \vec{r}')}{\lvert \vec{r} - \vec{r}'\rvert^3} B ( r ) = 4 π μ 0 ∮ L ∣ r − r ′ ∣ 3 I d ℓ ′ × ( r − r ′ ) Par a ˊ metros: r ⃗ = ( r , 0 , 0 ) , r ⃗ ′ = ( 0 , 0 , z ′ ) , d ℓ ⃗ ′ = z ^ d z ′ \text{Parámetros: } \vec{r}=(r,0,0),\; \vec{r}'=(0,0,z'),\; d\vec{\ell}'=\hat{z}\,dz' Par a ˊ metros: r = ( r , 0 , 0 ) , r ′ = ( 0 , 0 , z ′ ) , d ℓ ′ = z ^ d z ′ d ℓ ⃗ ′ × ( r ⃗ − r ⃗ ′ ) = r d z ′ y ^ , ∣ r ⃗ − r ⃗ ′ ∣ = r 2 + ( z ′ ) 2 \displaystyle d\vec{\ell}' \times (\vec{r}-\vec{r}')
= r\,dz'\;\hat{y}, \quad \lvert \vec{r} - \vec{r}'\rvert = \sqrt{r^2 + (z')^2} d ℓ ′ × ( r − r ′ ) = r d z ′ y ^ , ∣ r − r ′ ∣ = r 2 + ( z ′ ) 2 B ⃗ ( r ) = y ^ μ 0 I 4 π ∫ − ∞ + ∞ r d z ′ [ r 2 + ( z ′ ) 2 ] 3 / 2 = φ ^ μ 0 I 2 π r \displaystyle \vec{B}(r)
= \hat{y}\,\frac{\mu_0\,I}{4\pi} \int_{-\infty}^{+\infty} \frac{r\,dz'}{\bigl[r^2 + (z')^2\bigr]^{3/2}}
= \hat{\varphi}\,\frac{\mu_0\,I}{2\pi\,r} B ( r ) = y ^ 4 π μ 0 I ∫ − ∞ + ∞ [ r 2 + ( z ′ ) 2 ] 3/2 r d z ′ = φ ^ 2 π r μ 0 I Espira circular (radio R \Huge R R I \Huge I I Potencial vectorial A ⃗ ( r ⃗ ) = μ 0 4 π ∮ L I d ℓ ⃗ ′ ∣ r ⃗ − r ⃗ ′ ∣ \displaystyle \vec{A}(\vec{r}) = \frac{\mu_0}{4\pi} \oint_{L} \frac{I\,d\vec{\ell}'}{\lvert \vec{r} - \vec{r}'\rvert} A ( r ) = 4 π μ 0 ∮ L ∣ r − r ′ ∣ I d ℓ ′ Para r ⃗ = ( 0 , 0 , z ) , r ⃗ ′ = ( R cos φ ′ , R sin φ ′ , 0 ) , d ℓ ⃗ ′ = R d φ ′ φ ^ ′ \text{Para } \vec{r}=(0,0,z),\; \vec{r}'=(R\cos\varphi',\,R\sin\varphi',\,0),\; d\vec{\ell}'=R\,d\varphi'\,\hat{\varphi}' Para r = ( 0 , 0 , z ) , r ′ = ( R cos φ ′ , R sin φ ′ , 0 ) , d ℓ ′ = R d φ ′ φ ^ ′ A φ ( 0 , 0 , z ) = μ 0 I 4 π ∫ 0 2 π R d φ ′ R 2 + z 2 = μ 0 I R 2 R 2 + z 2 \displaystyle A_{\varphi}(0,0,z)
= \frac{\mu_0\,I}{4\pi} \int_{0}^{2\pi} \frac{R\,d\varphi'}{\sqrt{R^2 + z^2}}
= \frac{\mu_0\,I\,R}{2\,\sqrt{R^2+z^2}} A φ ( 0 , 0 , z ) = 4 π μ 0 I ∫ 0 2 π R 2 + z 2 R d φ ′ = 2 R 2 + z 2 μ 0 I R B ⃗ = ∇ × A ⃗ \Large \vec{B} = \nabla \times \vec{A} B = ∇ × A B z ( 0 , 0 , z ) = 1 r ∂ ∂ r [ r A φ ( r , z ) ] ∣ r = 0 = μ 0 I R 2 2 ( R 2 + z 2 ) 3 / 2 \displaystyle B_{z}(0,0,z)
= \frac{1}{r}\frac{\partial}{\partial r}\bigl[r\,A_{\varphi}(r,z)\bigr]\Big|_{r=0}
= \frac{\mu_0\,I\,R^2}{2\,(R^2 + z^2)^{3/2}} B z ( 0 , 0 , z ) = r 1 ∂ r ∂ [ r A φ ( r , z ) ] r = 0 = 2 ( R 2 + z 2 ) 3/2 μ 0 I R 2 Campo directo (Biot–Savart) Aplicamos directamente la ley de Biot–Savart para obtener B \mathbf{B} B
B ⃗ ( 0 , 0 , z ) = μ 0 I 4 π ∫ 0 2 π R d φ ′ ( φ ^ ′ × ( − z z ^ ) ) [ R 2 + z 2 ] 3 / 2 \displaystyle \vec{B}(0,0,z)
= \frac{\mu_0\,I}{4\pi} \int_{0}^{2\pi} \frac{R\,d\varphi'\;\bigl(\hat{\varphi}' \times (-\,z\,\hat{z})\bigr)}{\bigl[R^2 + z^2\bigr]^{3/2}} B ( 0 , 0 , z ) = 4 π μ 0 I ∫ 0 2 π [ R 2 + z 2 ] 3/2 R d φ ′ ( φ ^ ′ × ( − z z ^ ) ) Recordar que φ ^ ′ × z ^ = r ^ ′ , r ^ ′ \hat{\varphi}'\times\hat{z} = \hat{r}',\ \hat{r}' φ ^ ′ × z ^ = r ^ ′ , r ^ ′ B z B_z B z
B z ( 0 , 0 , z ) = μ 0 I R 2 2 ( R 2 + z 2 ) 3 / 2 \displaystyle B_{z}(0,0,z)
= \frac{\mu_0\,I\,R^2}{2\,\bigl(R^2 + z^2\bigr)^{3/2}} B z ( 0 , 0 , z ) = 2 ( R 2 + z 2 ) 3/2 μ 0 I R 2 Gracias a la simetría, solo la componente z z z φ \varphi φ
Simulador r → = ( 0.00 , 0.00 , 0.00 ) \overrightarrow{r} = (0.00, 0.00, 0.00) r = ( 0.00 , 0.00 , 0.00 ) B → ( r ) = ( 0.00 , 0.00 , 1.57 ) μ 0 4 π \overrightarrow{B}(r) = (0.00, 0.00, 1.57)\frac{\mu_0}{4\pi} B ( r ) = ( 0.00 , 0.00 , 1.57 ) 4 π μ 0 ∣ B → ( r ) ∣ = 1.5711 μ 0 4 π |\overrightarrow{B}(r)| = 1.5711 \frac{\mu_0}{4\pi} ∣ B ( r ) ∣ = 1.5711 4 π μ 0
