Question

A current carrying closed loop in the form of a right isosceles triangle \(XYZ\) is placed in a uniform magnetic field \(B\) acting along \(XY\) of the loop. If the magnetic force on the arm \(YZ\) is \(\sqrt{2}F\), then the force on the arm \(XZ\) is:

• A. \(-\sqrt{2}F\)
• B. \(\frac{F}{\sqrt{2}}\)
• C. \(+2\sqrt{2}F\)
• D. \(-F\)

Hint:

The net magnetic force on a closed current loop placed in a uniform magnetic field is always zero.

Answer 1

The Correct Option is A

Step 1: Use magnetic force on a current carrying wire.
\[ F = I l B \sin\theta \]
where \(\theta\) is the angle between length of conductor and magnetic field.

Step 2: Force on arm \(XY\).
Since magnetic field is along \(XY\),
\[ \theta = 0^\circ \]
\[ F_{XY} = I l B \sin 0^\circ = 0 \]

Step 3: Force on arm \(YZ\).
Arm \(YZ\) is perpendicular to \(XY\), so it is perpendicular to magnetic field.
\[ F_{YZ} = IB(YZ)\sin 90^\circ \]
Given,
\[ F_{YZ} = \sqrt{2}F \]

Step 4: Use net force condition on closed loop.
In a uniform magnetic field, the net magnetic force on a closed current loop is zero.
\[ F_{XY} + F_{YZ} + F_{XZ} = 0 \]

Step 5: Substitute known force.
\[ 0 + \sqrt{2}F + F_{XZ} = 0 \]

Step 6: Solve for force on \(XZ\).
\[ F_{XZ} = -\sqrt{2}F \]

Step 7: Final answer.
\[ \boxed{-\sqrt{2}F} \]