An element \(\Delta l=\Delta x\widehat i\) is placed at the origin and carries a large current \(I=10A\). What is the magnetic fleld on the \(y\)-axis at a distance of \(0.5\text{ m}.\Delta x=1\text{ cm}\)
Answer
\(\mid \mathit{dB}\mid =\frac{\mu _0}{4\pi }\frac{1\mathit{dl}\sin \theta }{r^2}\)
\(\mathit{dt}=\Delta x=10^{-2}m,l=10A,r=0.5m=y{\cdot}\mu _0\text /4\pi =10^{-7}\frac{\mathit{Tm}} A\)
\(\theta =90^{{\circ}}:\sin \theta =1\)
\(\left|\mathit{dB}\right|=\frac{10^{-7}\times 10\times 10^{-2}}{25\times 10^{-2}}=4\times 10^{-8}T\)
The direction of the field is in the \(+z\) -direction. This is so since. \(\mathit{dl}\times r=\Delta x\widehat i\times y\widehat j=y\Delta x\left(\widehat i\times \widehat j\right)=y\Delta x\widehat k\)
We remind you of the following cyclic property of cross-products. \(\widehat i\times \widehat j=\widehat k;\widehat j\times \widehat k=\widehat i;\widehat k\times \widehat i=\widehat j\)
Note that the field is small in magnitude.
\(\mathit{dt}=\Delta x=10^{-2}m,l=10A,r=0.5m=y{\cdot}\mu _0\text /4\pi =10^{-7}\frac{\mathit{Tm}} A\)
\(\theta =90^{{\circ}}:\sin \theta =1\)
\(\left|\mathit{dB}\right|=\frac{10^{-7}\times 10\times 10^{-2}}{25\times 10^{-2}}=4\times 10^{-8}T\)
The direction of the field is in the \(+z\) -direction. This is so since. \(\mathit{dl}\times r=\Delta x\widehat i\times y\widehat j=y\Delta x\left(\widehat i\times \widehat j\right)=y\Delta x\widehat k\)
We remind you of the following cyclic property of cross-products. \(\widehat i\times \widehat j=\widehat k;\widehat j\times \widehat k=\widehat i;\widehat k\times \widehat i=\widehat j\)
Note that the field is small in magnitude.
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