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Once My Wheel Gets Touched by the Power line , it Becomes a Rotating charged ring. In Other words , it's a current going around a loop ! To Find the magnitude of the current we note that my speed was 10 Km/H or $\frac{1000}{3.6}=277.8 M/S$ and my wheel was not slipping .Therefore the angular velocity of my wheel is $\omega$ = $\frac{v}{r}$ $\rightarrow$ $\frac{2.778}{0.3}= 9.26 Rad / s$ .

The Period of the Rotation is $T$ = $\frac{2\pi}{\omega} = 0.679 Seconds$ . Now , How does this give the current in a Loop ? Imagine you sit at a point at the lop of the wheel but don't spin around with it . 1C of charge passes by you every $0.679$ .Therefore the current is $\frac{1}{0.679} = 1.473 A$ . The Magnitude of the magnetic field a distance z from the center along the axis of rotation of a Loop of Radius R carrying current I Is

$B= \displaystyle \frac{\mu_o}{2}$ $\frac{ I R^2}{(z^2 + R^2)^\frac{3}{2}}$

Substituting numbers yields B = $\displaystyle2.63 \times 10^{-6}$ or $0.00000263$

$Q$ chaged ring rotating with $w$ angular velocity is equivalent to a loop having $i = \frac{Qw}{2 \pi}$ .

Proof: Let the frequency be $f$ per second, $f$ revolutions take place and hence $fQ$ charge passes through a cross-section , Hence $i = fQ = \frac{Qw}{2 \pi}$

At a distance $x$ from the center of a circular loop on its axis ,

$B = \frac{\mu_{0}iR^2}{2(x^2 + R^2)^{\frac{3}{2}}}$

= $\frac{\mu_{0}QwR^2}{4 \pi (x^2 + R^2)^{\frac{3}{2}}} = \boxed{2.63 \times 10^{-6}T}$ ( $w = \frac{v}{r} = 9.26 rad/sec.$ )

Since it's given that

$v=10km/h=2.778m/s$

$T=\frac{2\pi R}{v}$

therefore

$I=\frac{Q}{T}=\frac{Qv}{2\pi R}$

The magnetic field at a horizontal distance $x$ from the center of the current loop is

$B_{x}=\int \frac{\mu_{0}I}{4\pi}\frac{Rdl}{(x^2+R^2)^{3/2}}=\frac{\mu_{0}IR}{4\pi (x^2+R^2)^{3/2}} \int dl$

since $\int dl = 2\pi R$

Substituting, we get

$B= \frac{\mu_{0}QvR}{4\pi (x^2+R^2)^{3/2}}= 2.63\times 10^{-6} T$

I didn't understand properly why we neglected the translational velocity of the ring. I thought it is neglected as the observer is sitting on the bike , but does this imply that the electromagnetic fields are relative? Sorry to ask such a silly question but I didn't it understand properly. Thanks in advance.

Once My Wheel Gets Touched by the Power line , it Becomes a Rotating charged ring. In Other words , it's a current going around a loop ! To Find the magnitude of the current we note that my speed was 10 Km/H or and my wheel was not slipping .Therefore the angular velocity of my wheel is = . The Period of the Rotation is = . Now , How does this give the current in a Loop ? Imagine you sit at a point at the lop of the wheel but don't spin around with it . 1C of charge passes by you every .Therefore the current is . The Magnitude of the magnetic field a distance z from the center along the axis of rotation of a Loop of Radius R carrying current I Is

Substituting numbers yields B = or Reply Subscribe to comments

Considering the wheel as a loop, the rotation of the wheel leads to the charge thus constituting current. So, first find the current due to the rotation of the loop,

$I = \frac {Q}{T}$ here, $Q = 1C$ , $T=$ Time period of rotation i.e

$T=\frac { 2\pi r}{v} = \frac { 2 \pi \times 0.3}{10 \times \frac {5}{18} } = 0.6785 s$

$I = \frac{1}{0.67858} = 1.4736 A$ ......(1)

Now, calculating the Magnetic field at a distance $d(0.1 m)$ perpendicular from the centre of the front wheel. This situation is similar to finding the magnetic field at a point on the axis of the loop. Which is given by

$B = \frac { \mu_{0} 2 \pi r^{2} I}{4 \pi (r^{2} + d^{2})^{\frac {3}{2} } }$

On substituting the values we get, $B = 0.00000263$ T or $B = 2.63E-6$ Tesla .