Question

The force acting per unit length when a very long straight conductor is carrying a steady current of 1 A and the direction of the current is from south to north is (The horizontal component of the earth's magnetic field at the place is \( 3 \times 10^{-5} \) T and the direction of the field is from the geographical south to geographical north.)

• A. \( 3 \times 10^{-5} \) Nm\(^{-1} \)
• B. \( 1 \times 10^{-5} \) Nm\(^{-1} \)
• C. \( 0 \)
• D. \( 1.5 \times 10^{-5} \) Nm\(^{-1} \)

Hint:

The force on a current-carrying conductor in a magnetic field depends on the angle between the current direction and the magnetic field direction. If the current is parallel or anti-parallel to the magnetic field, the force is zero because \( \sin(0^\circ) = 0 \) and \( \sin(180^\circ) = 0 \).

Answer 1

The Correct Option is C

The force per unit length on a straight current-carrying conductor in a magnetic field is given by the formula: $$ \vec{F}/l = \vec{I} \times \vec{B} $$ where \( \vec{F}/l \) is the force per unit length, \( \vec{I} \) is the current vector (magnitude is the current and direction is the direction of the current), and \( \vec{B} \) is the magnetic field vector.
The magnitude of the force per unit length is: $$ |\vec{F}/l| = I B \sin \theta $$ where \( I \) is the current, \( B \) is the magnitude of the magnetic field, and \( \theta \) is the angle between the direction of the current and the direction of the magnetic field.
In this problem: Current \( I = 1 \) A, direction is from geographical south to geographical north.
Horizontal component of the Earth's magnetic field \( B = 3 \times 10^{-5} \) T, direction is from geographical south to geographical north.
The direction of the current is the same as the direction of the magnetic field.
Therefore, the angle \( \theta \) between the current vector and the magnetic field vector is \( 0^\circ \).
The force per unit length is: $$ |\vec{F}/l| = (1 \text{ A}) \times (3 \times 10^{-5} \text{ T}) \times \sin(0^\circ) $$ Since \( \sin(0^\circ) = 0 \), the force per unit length is: $$ |\vec{F}/l| = 3 \times 10^{-5} \times 0 = 0 $$ The force acting per unit length on the conductor is zero.