Question

A current carrying conductor is placed in an uniform magnetic field. The direction of current is parallel to the direction of magnetic field. The magnetic force $F$ acting on the conductor is_______. ( Length of the conductor $L$, current $I$).

• A. zero
• B. $\text{F}=\text{ILB}$
• C. $\text{F}=\text{IL}{\text{B}}^2$
• D. $\text{F}$ = $\text{ILB/2}$

Answer 1

The correct option is A zero

Let $\theta$ be the angle between the current carrying wire of length L and the magnetic field B. If I is the current then the force F acting on the wire due to the magnetic field is given by

$\text{F}=\text{ILB(sin}\theta )$

The wire is placed in the same direction of magnetic field, i.e. the angle between the current and the field is $0^{\circ }$. So, $\text{F}=\text{ILB(sin}{0}^{\circ })$ = $0$, since sin $0^{\circ }$ = $0$