Question

The horizontal component of the Earth's magnetic field at a certain place is $3\times {10}^{-5}T$ 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 $1A.$ The force per unit length on it, when it is placed on a horizontal table and the direction of the current is east to west is :

• A. $3\times {10}^{-5}N{m}^{-1}$
• B. $6\times {10}^{-3}N{m}^{-1}$
• C. $9\times {10}^{-2}N{m}^{-1}$
• D. $12\times {10}^{-6}N{m}^{-1}$

Answer 1

The correct option is A $3\times {10}^{-5}N{m}^{-1}$

As, $F=I\overrightarrow{l}\times \overrightarrow{B}$ ,
$F=IIB\sin \theta$
The force per unit length is,
$f=\frac{F}{l}=IB\sin \theta$
When the current is flowing from east to west,
$\theta ={90}^o,$ hence,
$f=IB=1\times 3\times {10}^{-5}=3\times {10}^{-5}N{m}^{-1}$