Verified Solution • 2025–26 Session moderate

QPages 230-234: Q12: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q12 Solution: A circular loop of radius , carrying a current , is placed in a …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume that because the magnetic field is non-uniform or radial, a net force must exist; however, the local force $d\vec{F} = i d\vec{l} \times \vec{B}$ is radial at every point, and these opposing radial components cancel out perfectly due to the loop's symmetry, resulting in zero net force.

A circular loop of radius

a

, carrying a current

i

, is placed in a two-dimensional magnetic field. The centre of the loop coincides with the centre of the field figure). The strength of the magnetic field at the periphery of the loop is B. Find the magnetic force on the wire.

Step-by-Step Verified Solution

**Answer:**

F = 0

. **Step-by-Step Explanation:** The magnetic field is radial (directed towards or away from the center). The current element

d\vec{l}

is tangential to the loop. Since

d\vec{l}

is perpendicular to

\vec{B}

at every point, the force

d\vec{F} = i d\vec{l} \times \vec{B}

is directed radially. By symmetry, the net force on the entire loop is zero.

All Questions in this Chapter

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