Referring to the loop ABCD in the previous question, a straight thin wire with steady current I_1 flowing out of the plane of the paper is kept at the - Physics MCQ AIEEE 2009 2026 | ExamTest.live

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Referring to the loop A BC D in the previous question, a straight thin wire with steady current I_{1} flowing out of the plane of the paper is kept at the origin. Due to the presence of the current I_{1} at the origin:

A $The forces on A B and D C are zero.$
B $The forces on A D and BC are zero.$
C $The magnitude of the net force on the loop is \frac{μ _{0} I I _{1}}{4 π} \frac{2 π}{3} \frac{b - a}{a + b} .$
D $The magnitude of the net force on the loop is \frac{μ _{0} I I _{1} ( b - a )}{24 ab} .$

Solution & Step-by-step Explanation

The magnetic field B_{1} produced by the wire at the origin is circular. At any point on the arcs A D and BC, the magnetic field vector is tangent to the arc. The current element I d l of the loop along the arcs is also tangent to the arc.Since d l ∥ B_{1}, the force d F = I (d l \times B_{1}) = 0 .Thus, the forces on the arcs A D and BC are zero.

Try it yourself before checking the explanation above.

Referring to the loop A BC D in the previous question, a straight thin wire with steady current I_{1} flowing out of the plane of the paper is kept at the origin. Due to the presence of the current I_{1} at the origin:

The forces on A B and D C are zero.

The forces on A D and BC are zero.

The magnitude of the net force on the loop is \frac{μ _{0} I I _{1}}{4 π} \frac{2 π}{3} \frac{b - a}{a + b} .

The magnitude of the net force on the loop is \frac{μ _{0} I I _{1} ( b - a )}{24 ab} .

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