2011 AIPMT-PRE Moving Charges and MagnetismForce on a Current-carrying Conductor in a Magnetic Field Medium Question 5 of 7 in Force on a Current-carrying Conductor in a Magnetic Field

A current carrying closed loop in the form of a right angle isosceles triangle ABC is placed in a uniform…

A current carrying closed loop in the form of a right angle isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is $\vec{F}$, the force on the arm AC is
A $\sqrt{2}\vec{F}$
B $-\sqrt{2}\vec{F}$
C $-\vec{F}$
D $\vec{F}$

The answer is C.

Solution

Force on a current-carrying straight conductor: $\vec{F} = I\vec{L}\times\vec{B}$
The net force on a closed current loop in a uniform magnetic field is zero: $\vec{F}_{AB} + \vec{F}_{BC} + \vec{F}_{CA} = 0$
The field is along AB, so for the arm AB the length vector is parallel to $\vec{B}$ and $\vec{F}_{AB} = 0$.
Therefore $\vec{F}_{BC} + \vec{F}_{CA} = 0$
$\vec{F}_{AC} = -\vec{F}_{BC} = -\vec{F}$

Concept Force on a closed loop

Force on a Current-carrying Conductor in a Magnetic Field in past papers

7 questions from this chapter have appeared across 6 exam years.