When a wire is moved inside uniform magnetic field then its free electrons will experience magnetic force on it due to which wire will have potential difference at its ends.
Now here we will have magnetic field due to earth and wire is moving in this constant field so induced emf is given by formula
[tex]EMF = v.(B x L)[/tex]
given that
[tex]B = 25\mu Tj - 50\mu Tk[/tex]
[tex] v = 2 m/s j[/tex]
[tex]L = 0.50 m (-i)[/tex]
now by using the above formula we will have
[tex]EMF = 2(j) .(25\mu j - 50\mu k) x (-0.50 i)[/tex]
[tex]EMF = 2(j) .(12.5\mu k + 25\mu j) [/tex]
[tex]EMF = 50 \mu Volts[/tex]
The emf induced between the ends of the wire is about 50 μV
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Let's recall the induced emf formula for straight wire moving in magnetic field as follows:
[tex]\boxed {\varepsilon = B L v}[/tex]
where:
ε = induced emf ( V )
B = magnetic field strength ( T )
L = length of wire ( m )
v = speed of wire ( m/s )
B, L , v are mutually perpendicular
Let us now tackle the problem!
Given:
speed of wire = v = 2.0 m/s
length of wire = L = 50 cm
magnetic field strength = B = 50 μT → vertical component
Asked:
induced emf = ε = ?
Solution:
[tex]\varepsilon = B L v[/tex]
[tex]\varepsilon = (50\ \mu T)(50 \ cm)(2.0 \ m/s)[/tex]
[tex]\varepsilon = (50\ \mu T)(0.5 \ m)(2.0 \ m/s)[/tex]
[tex]\varepsilon = (50\ \mu T)(1.0 \ m^2/s)[/tex]
[tex]\varepsilon = 50\ \mu V[/tex]
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The emf induced between the ends of the wire is about 50 μV
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Grade: High School
Subject: Physics
Chapter: Electromagnetism
