Question

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. $-\sqrt2\, \vec{F}$
• B. $-\vec{F}$
• C. $\vec{F}$
• D. $\sqrt2 \,\vec{F}$

Answer 1

The Correct Option is B

To solve the problem, we need to determine the magnetic force on the arm $AC$ when the given current-carrying loop is subjected to a uniform magnetic field acting along $AB$.

The setup describes a right angle isosceles triangle, implying that the sides $AB$ and $AC$ are equal, and side $BC$ is the hypotenuse.

Key physics concepts involved in this problem include:

• The Lorentz force law, which describes the force on a current-carrying conductor in a magnetic field: $F = I \cdot (L \times B)$, where $I$ is the current, $L$ is the length vector of the conductor, and $B$ is the magnetic field.
• In a uniform magnetic field, the forces on the current segments $AB$ and $AC$ will be along directions perpendicular to them.

Since the magnetic field is along $AB$, there is no force on $AB$ (because the angle between $L$ and $B$ is zero and $\sin 0 = 0$). However, forces will act on $AC$ and $BC$.

Force on Arm $BC$: The problem states that the force on $BC$ is $\vec{F}$.

Force on Arm $AC$: As $BC$ and $AC$ are the arms of the triangle, they are perpendicular to each other. By symmetry and using the equilibrium condition (since a current loop in a uniform magnetic field will attempt to align itself such that the net force is zero), the force on $AC$ will be equal in magnitude and opposite in direction to the force on $BC$.

Thus, the force on $AC$ is -\vec{F}.

The correct answer is: -\vec{F}.