The horizontal component of the earth's magnetic field at a certain place is $3.0 \times 10^{-5}\; T$ and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of $1 \;A$. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is $(a)$ east to west; $(b)$ south to north?
The horizontal component of the earth's magnetic field at a certain place is $3.0 \times 10^{-5}\; T$ and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of $1 \;A$. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is $(a)$ east to west; $(b)$ south to north?
$F =I l \times B$
$F=I L B \sin \theta$
The force per unit length is
$f=F / l=I B \sin \theta$
$(a)$ When the current is flowing from east to west. $\theta=90^{\circ}$
Hence, $f=I B$
$=1 \times 3 \times 10^{-5}=3 \times 10^{-5} N m ^{-1}$
This is larger than the value $2 \times 10^{-7} Nm ^{-1}$ quoted in the definition of the ampere. Hence it is important to eliminate the effect of the earth's magnetic field and other stray fields while standardising the ampere.
The direction of the force is downwards. This direction may be obtained by the directional property of cross product of vectors.
$(b)$ When the current is flowing from south to north, $\theta=0^{\circ}$
$f=0$
Hence there is no force on the conductor
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