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For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by, \frac{\mu _0 I R^2 N}{2(x^2 + R^2)^{\frac{3}{2}}} (a) Show that this reduces to the familiar result for field at the centre of the coil.
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Anonym0us
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Qalaxia Knowlege Bot
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I found an answer from en.wikipedia.org

Magnetic field - Wikipedia


A magnetic field is a vector field that describes the magnetic influence on moving electric charges, electric currents, :ch1 and magnetized materials. A charge that ...


For more information, see Magnetic field - Wikipedia

Qalaxia Master Bot
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I found an answer from www.khanacademy.org

Introduction to magnetism (video) | Khan Academy


An introduction to magnetism. ... Magnetic force on a current carrying wire ... When a bar magnet is broken in the middle, what is the effect on its intensity? ... In this time, the Earth's magnetic field automatically aligned the domains, or atoms, of the ... charge) or changing electric fields and create force on moving charges only.


For more information, see Introduction to magnetism (video) | Khan Academy

Guru Nanak
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Given that

Coil radius = R

Number of turns = N

Current = I

Distance between  magnetic field and centre = x


(a) Show that this reduces to the familiar result for field at the centre of the coil.

        Magnetic field B = \frac{\mu _0 I R^2 N}{2(x^2 + R^2)^{\frac{3}{2}}}   


        Magnetic field at centre x = 0

            B=\frac{\mu_0IR^2N}{2\left(0^2+\ R^2\right)^{\frac{3}{2}}}  


             B = \frac{\mu _0 I R^2 N}{2R^3}   


             B = \frac{\mu _0 I N}{2R}   


        Hence, B reduces to the familiar result for field at the centre of the coil