The fractional change in the magnetic field i | vedclass

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(D) The magnetic field on the axis of a current-carrying coil of radius $a$ at a distance $r$ from the centre is given by $B_{	ext{axis}} = \frac{\mu

Vedclass · Accepted Answer

$\frac{3}{2} \frac{r^{2}}{a^{2}}$

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(D) The magnetic field on the axis of a current-carrying coil of radius $a$ at a distance $r$ from the centre is given by $B_{	ext{axis}} = \frac{\mu_{0} i a^{2}}{2(a^{2} + r^{2})^{3/2}}$.The magnetic field at the centre of the coil is $B_{	ext{centre}} = \frac{\mu_{0} i}{2a}$.The fractional change in magnetic field is defined as $\frac{B_{	ext{centre}} - B_{	ext{axis}}}{B_{	ext{centre}}} = 1 - \frac{B_{	ext{axis}}}{B_{	ext{centre}}}$.Substituting the expressions: $1 - \frac{\frac{\mu_{0} i a^{2}}{2(a^{2} + r^{2})^{3/2}}}{\frac{\mu_{0} i}{2a}} = 1 - \frac{a^{3}}{(a^{2} + r^{2})^{3/2}} = 1 - \left(1 + \frac{r^{2}}{a^{2}}ight)^{-3/2}$.Using the binomial approximation $(1 + x)^{n} \approx 1 + nx$ for $x $1 - (1 - \frac{3}{2} \frac{r^{2}}{a^{2}}) = \frac{3}{2} \frac{r^{2}}{a^{2}}$.

The fractional change in the magnetic field intensity at a distance $r$ from the centre on the axis of a current-carrying coil of radius $a$ to the magnetic field intensity at the centre of the same coil is: (Take $r << a$)

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$A$ straight section $PQ$ of a circuit lies along the $X$-axis from $x = -\frac{a}{2}$ to $x = \frac{a}{2}$ and carries a steady current $i$. The magnetic field due to the section $PQ$ at a point $x = a$ on the $X$-axis will be:

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