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Question

PQ is a long straight conductor carrying a current of $$3$$A as shown in Figure. An electron moves with a velocity of $$2\times 10^7 ms^{-1}$$ parallel to it. Find the force acting on the electron.
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Solution

Magnetic field at a perpendicular distance $$d$$ due to long straight conductor is given by  $$B = \dfrac{\mu_o I}{2\pi d}$$
Given : $$I = 3 \ A$$,  $$d = 0.6 \ m$$
We know $$\mu_o = 4\pi \times 10^{-7}$$ $$T m A^{-1}$$
$$\implies$$  $$B = \dfrac{2\times 10^{-7} \times 3 }{ (0.6)} = 10^{-6}  \ T$$
Speed of the electron  $$v = 2\times 10^7  \ m/s$$
$$\therefore$$ Force acting on the electron  $$|F| = |q| vB$$
$$\therefore$$  Force acting on the electron  $$|F| = (1.6\times 10^{-19})(2\times 10^7)(10^{-6}) = 3.2\times 10^{-18}  \ N$$

Physics

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