0.4 Magnetic field due to current in a circular wire  (Page 4/5)

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$R=\frac{L}{2\pi }$

The magnetic field due to current I in the circular wire is :

${B}_{C}=\frac{{\mu }_{0}I}{2R}=\frac{2\pi {\mu }_{0}I}{2L}=\frac{\pi {\mu }_{0}I}{L}=\frac{3.14{\mu }_{0}I}{L}$

In the case of straight wire, let us consider that wire is long enough for a point around middle of the wire. For comparison purpose, we assume that perpendicular linear distance used for calculating magnetic field due to current in straight wire is equal to the radius of circle. The magnetic field at a perpendicular distance “R” due to current in long straight wire is given as :

${B}_{L}=\frac{{\mu }_{0}I}{4\pi R}=\frac{{\mu }_{0}IX2\pi }{4\pi L}=\frac{{\mu }_{0}I}{2L}=\frac{0.5{\mu }_{0}I}{L}$

Clearly, the magnetic field due to current in circular wire is 6.28 times greater than that due to current in straight wire at comparable points of observations. Note that this is so even though we have given advantage to straight wire configuration by assuming it to be long wire. In a nutshell, a circular configuration tends to concentrate magnetic field along axial direction which is otherwise spread over the whole length of wire.

Problem : A current 10 A flowing through a straight wire is split at point A in two semicircular wires of radius 0.1 m. The resistances of upper and lower semicircular wires are 10 Ω and 20 Ω respectively. The currents rejoin to flow in the straight wire again as shown in the figure. Determine the magnetic field at the center “O”.

Solution : The straight wire sections on extension pass through the center. Hence, magnetic field due to straight wires is zero. Here, the incoming current at A is distributed in the inverse proportion of resistances. Let the subscripts “1” and “2’ denote upper and lower semicircular sections respectively. The two sections are equivalent to two resistances in parallel combination as shown in the figure. Here, potential difference between “A” and “B” is :

${V}_{AB}=\frac{IX{R}_{1}X{R}_{2}}{\left({R}_{1}+{R}_{2}\right)}={I}_{1}{R}_{1}={I}_{2}{R}_{2}$ $⇒{I}_{1}=\frac{IX{R}_{2}}{\left({R}_{1}+{R}_{2}\right)}=\frac{10X20}{30}=\frac{20}{3}\phantom{\rule{1em}{0ex}}A$ $⇒{I}_{2}=\frac{IX{R}_{1}}{\left({R}_{1}+{R}_{2}\right)}=\frac{10X10}{30}=\frac{10}{3}\phantom{\rule{1em}{0ex}}A$

We see that current in the upper section is twice that in the lower section i.e. ${I}_{1}=2{I}_{2}$ . Also, the magnetic field is perpendicular to the plane of semicircular section (plane of drawing). The current in the upper semicircular wire is clockwise. Thus, the magnetic field due to upper section is into the plane of drawing. However, the current in the lower semicircular is anticlockwise. Thus, the magnetic field due to lower section is out of the plane of drawing. Putting θ = π for each semicircular section, the net magnetic field due to semicircular sections at “O” is:

$B=\frac{{\mu }_{0}{I}_{1}\pi }{4\pi R}-\frac{{\mu }_{0}{I}_{2}\pi }{4\pi R}$ $⇒B=\frac{{\mu }_{0}{I}_{1}\pi }{4\pi R}-\frac{{\mu }_{0}{I}_{1}\pi }{8\pi R}=\frac{{\mu }_{0}{I}_{1}\pi }{8\pi R}$ $⇒B=\frac{{10}^{-7}X20}{3X8X0.1}=8.3X{10}^{-7}\phantom{\rule{1em}{0ex}}T$

The net magnetic field is into the plane of drawing.

Exercises

An electron circles a single proton nucleus of radius $3.2X{10}^{-11}$ m with a frequency of ${10}^{16}$ Hz. The charge on the electron is $1.6X{10}^{-19}$ Coulomb. What is the magnitude of magnetic field due to orbiting electron at the nucleus?

The equivalent current is given by :

$I=\frac{q}{T}=q\nu$

where ν and T are frequency and time period of revolutions respectively. The magnitude of magnetic field due to circular wire is given by :

$B=\frac{{\mu }_{0}I}{2R}$

Substituting for I, we have :

$⇒B=\frac{{\mu }_{0}I}{2R}=\frac{{\mu }_{0}q\nu }{2R}$

Putting values,

$⇒B=\frac{4\pi {10}^{-7}X1.6X{10}^{-19}X{10}^{16}}{2X3.2X{10}^{-11}}$ $⇒B=31.4\phantom{\rule{1em}{0ex}}T$

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