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Exercises · 6.6

Q.A horizontal straight wire 10 m10\ \text{m} long extending from east to west is falling with a speed of 5.0 m s−15.0\ \text{m s}^{-1}, at right angles to the horizontal component of the earth's magnetic field, 0.30×10−4 Wb m−20.30 \times 10^{-4}\ \text{Wb m}^{-2}.

(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
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Motional emf E=BH l v=1.5×10−3 V\mathcal{E}=B_H\,l\,v = 1.5\times10^{-3}\ \text{V}; the emf drives positive charge from west to east, so the east end is at the higher potential.

The falling wire cuts the horizontal component of Earth's magnetic field, inducing a motional emf E=BH l v\mathcal{E}=B_H\,l\,v — valid because the wire, its velocity, and the field are mutually perpendicular.

  1. Magnitude of the induced emf With BH=0.30×10−4 T=3.0×10−5 TB_H = 0.30\times10^{-4}\ \text{T}=3.0\times10^{-5}\ \text{T}, l=10 ml=10\ \text{m}, v=5.0 m s−1v=5.0\ \text{m s}^{-1}:

    E=BH l v=(3.0×10−5)(10)(5.0)=1.5×10−3 V=1.5 mV.\mathcal{E}=B_H\,l\,v=(3.0\times10^{-5})(10)(5.0)=1.5\times10^{-3}\ \text{V}=1.5\ \text{mV}.

  2. Direction of the emf Take east =i^=\hat{i}, north =j^=\hat{j}, up =k^=\hat{k}. The velocity is v⃗=−vk^\vec v=-v\hat k (downward) and the horizontal field points north, B⃗=BHj^\vec B=B_H\hat j. The force per unit charge on a positive carrier is …

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