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Home » For a circular coil of radius and turns carrying current , the magnitude of the magnetic field at a point on its axis at a distance from its centre is given by, (a) Show that this reduces to the familiar result for field at the centre of the coil. (b) Consider two parallel co-axial circular coils of equal radius , and number of turns , carrying equal currents in the same direction, and separated by a distance Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to , and is given by, , approximately.
For a circular coil of radius \mathrm{R} and \mathrm{N} turns carrying current \mathrm{I}, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by, B=\frac{\mu_{0} I R^{2} N}{2\left(x^{2}+R^{2}\right)^{3 / 2}} (a) Show that this reduces to the familiar result for field at the centre of the coil. (b) Consider two parallel co-axial circular coils of equal radius \mathbf{R}, and number of turns \mathbf{N}, carrying equal currents in the same direction, and separated by a distance \mathbf{R} . Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to \mathbf{R}, and is given by, B=0.72 \frac{\mu_{0} N I}{R}, approximately.
CBSE | Class 12 | Moving Charges and Magnetism | NCERT | Physics
For a circular coil of radius \mathrm{R} and \mathrm{N} turns carrying current \mathrm{I}, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by, B=\frac{\mu_{0} I R^{2} N}{2\left(x^{2}+R^{2}\right)^{3 / 2}} (a) Show that this reduces to the familiar result for field at the centre of the coil. (b) Consider two parallel co-axial circular coils of equal radius \mathbf{R}, and number of turns \mathbf{N}, carrying equal currents in the same direction, and separated by a distance \mathbf{R} . Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to \mathbf{R}, and is given by, B=0.72 \frac{\mu_{0} N I}{R}, approximately.

(a) Given is the expression of B as \frac{\mu_{0} I R^{2} N}{2\left(x^{2}+R^{2}\right)^{3 / 2}}

x will be zero at the centre of the coil,

Therefore, the magnetic field at the centre is B=\frac{\mu_{0} I R^{2} N}{2 R^{3}}

B=\frac{\mu_{0} I N}{2 R}

(b) Let d be the small distance between the point P and O.

Distance for the coil C will be O_{1} P=x_{1}=(R / 2)+d

Distance for the coil d will be

O_{2}P=x_{2}=(\mathrm{R} / 2)-\mathrm{d}

The magnetic field on the axis at a distance x from the centre of a circular coil with a radius of a, n turns, and a current I is given by

B=\frac{\mu_{0}}{4 \pi} \frac{2 \pi n I a^{2}}{\left(x^{2}+a^{2}\right)^{3 / 2}}

B=\frac{\mu_{0}}{2} \frac{n I a^{2}}{\left(x^{2}+a^{2}\right)^{3 / 2}}

The magnetic field at P due to coil C

B_{1}=\frac{\mu_{0}}{2} \frac{N I R^{2}}{\left(x_{1}^{2}+R^{2}\right)^{3 / 2}}

B_{1}=\frac{\mu_{0}}{2} \frac{N I R^{2}}{\left(\left(\frac{R}{2}+d\right)^{2}+R^{2}\right)^{3 / 2}}

B_{1}=\frac{\mu_{0}}{2} \frac{N I R^{2}}{\left(\left(\frac{R^{2}}{4}+d^{2}+R d\right)+R^{2}\right)^{3 / 2}}

\mathrm{d}^{2} can be neglected when compared to \mathrm{R}^{2}

B_{1}=\frac{\mu_{0} N I R^{2}}{2} \frac{1}{\left(\frac{5 R^{2}}{4}\right)^{3 / 2}}\left[\left(1+\frac{4 d}{5 R}\right)^{-3 / 2}\right]

It acts along \mathrm{PO}_{2}

The magnetic field at \mathrm{P} given by

B_{2}=\frac{\mu_{0}}{2} \frac{N I a^{2}}{\left(x_{2}^{2}+R^{2}\right)^{3 / 2}}

B_{2}=\frac{\mu 0}{2} \frac{N I R^{2}}{\left(\left(\frac{R}{2}-d\right)^{2}+R^{2}\right)^{3 / 2}}

B_{2}=\frac{\mu_{0} N I R^{2}}{2} \frac{1}{\left(\frac{B R^{2}}{4}\right)^{3 / 2}}\left[\left(1-\frac{4 d}{5 R}\right)^{-3 / 2}\right]

It acts along \mathrm{P} \mathrm{O}_{2}

The total Magnetic field at the point P due to the current through the coils B=B_{1}+B_{2}

B=\frac{\mu_{0} N I R^{2}}{2} \frac{1}{\left(\frac{R^{2}}{4}\right)^{3 / 2}}\left[\left(1+\frac{4 d}{5 R}\right)^{-3 / 2}+\left(1-\frac{4 d}{5 R}\right)^{-3 / 2}\right]

B=\frac{\mu_{0} N I R^{2}}{2\left(\frac{5, R^{2}}{4}\right)^{3 / 2}}\left[1-\frac{6 d}{5 R}+1+\frac{6 d}{5 R}\right]

B=\frac{\mu_{0} N I R^{2}}{2\left(\frac{M \pi^{2}}{4}\right)^{3 / 2}} \times 2

B=\frac{\mu_{0} N I}{R}\left(\frac{4}{5}\right)^{3 / 2}

B=(0.72) \frac{\mu_{0} N I}{R}

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