A long straight wire carries a current $$I_{0}$$. At distances $$a$$ and $$b$$ from it there are two other wires, parallel to the former one, which are interconnected by a resistance $$R$$ (Fig.). A connector slides without friction along the wires with a constant velocity $$v$$. Assuming the resistances of the wires, the connector, the sliding contacts, and the self-inductance of the frame to be negligible, find: (a) the magnitude and the direction of the current induced in the connector; (b) the force required to maintain the connector's velocity constant.
As $$\vec {B}$$, due to the straight current carrying wire, varies along the rod (connector) and enters linerarly so, to make the calculations simple, $$\vec {B}$$ is made constant by taking its average value in the range $$[a, b]$$.
$$< B > = \dfrac {\int_{a}^{b} B\ dr}{\int_{a}^{b} dr} = \dfrac {\int_{a}^{b} \dfrac {\mu_{0}}{2\pi}\dfrac {i_{0}}{r}dr}{\int_{a}^{b} dr}$$
or, $$< B > = \dfrac {\mu_{0}}{2\pi} \dfrac {i_{0}}{(b - a)} ln \dfrac {b}{a}$$
(a) The flux of $$\vec {B}$$ changes through the loop due to the movement of the connector. According to Lenz's law, the current in the loop will be anticlockwise. The magnitude of motional e.m.f.,
$$\xi_{in} = v < B > (b - a)$$
$$= \dfrac {\mu_{0}}{2\pi} \dfrac {i_{0}}{(b - a)} ln \dfrac {b}{a} (b - a) \dfrac {dx}{dt} = \dfrac {\mu_{0}}{2\pi} i_{0} ln \dfrac {b}{a}v$$
So, induced current
$$i_{in} = \dfrac {\xi_{in}}{R} = \dfrac {\mu_{0}}{2\pi} \dfrac {i_{0}v}{R} ln \dfrac {b}{a}$$
(b) The force required to maintain the constant velocity of the connector must be the magnitude equal to that of Ampere's acting on the connector, but in opposite direction.
So, $$F_{ext} = i_{in}l < B > = \left (\dfrac {\mu_{0}}{2\pi} \dfrac {i_{0}}{R} v\ln \dfrac {b}{a}\right ) (b - a) \left (\dfrac {\mu_{0}}{2\pi} \dfrac {i_{0}}(b - a)} ln \dfrac {b}{a}\right )$$
$$= \dfrac {v}{R} \left (\dfrac {\mu_{0}}{2\pi} i_{0} ln \dfrac {b}{a}\right )^{2}$$, and will be directed as shown in the (Fig.)
The figure shows a part of an electric circuit. The wires AB,CD and EF are long and have identical resistances. The separation between the neighboring wires is 1.0cm . The wires EA and BF have negligible resistance and the ammeter reads 30 A. Calculate the magnetic force per unit length on AB and CD.
A beam of electrons is moving with constant velocity in a region having simultaneous perpendicular electric and magnetic fields of strength 20V/m and 0.5T respectively at right angles to the direction of motion of the electrons. Then the velocity of electrons must be
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The direction of induced potential difference is given by :
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State the explain Fleming's right hand rule.
State Fleming's right hand rule.
The direction of induced current is obtained by:
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