CBSE-XI-Physics
35: Magnetic Field due to a Current
- #1The magnetic field at the origin due to a current element
i dl→placed at a position
r→is
(a)
μ0i4πdl →×r →r3
(b)
-μ0i4πr →×dl →r3
(c)
μ0i4πr →×dl →r3
(d)
-μ0i4πdl →×r →r3digAnsr: a,bAns : (a) `` \frac{{\,\mathrm{\,\mu \,}}_{0}i}{4\pi }\frac{d\stackrel{\to }{l}\times \stackrel{\to }{r}}{{r}^{3}}``
(b) `` -\frac{{\,\mathrm{\,\mu \,}}_{0}i}{4\pi }\frac{\stackrel{\to }{r}\times d\stackrel{\to }{l}}{{r}^{3}}``
The magnetic field at the origin due to current element `` id\stackrel{\to }{l}`` placed at a position `` \stackrel{\to }{r}`` is given by
`` d\stackrel{\to }{B}=\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}}{4\,\mathrm{\,\pi \,}}i\frac{\stackrel{\to }{r}\times d\stackrel{\to }{l}}{{r}^{3}}``
`` ``
According to the cross product property,
`` \stackrel{\to }{A}\times \stackrel{\to }{B}=-\stackrel{\to }{B}\times \stackrel{\to }{A}``
`` \Rightarrow d\stackrel{\to }{B}=-\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}}{4\,\mathrm{\,\pi \,}}i\frac{\stackrel{\to }{r}\times \stackrel{\to }{dl}}{{r}^{3}}``
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