Verified Solution • 2025–26 Session moderate

QPages 249-253: Q40: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q40 Solution: A circular loop of radius carrying a current is held at the…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often equate the torque on the small loop directly with the required holding force, forgetting that the force is applied at a distance $r$ from the axis. The correct relation is $F= au/r$, so the force scales with $r$, not $r^2$. Ensure you use the proper lever arm when converting torque to force, and check that the final expression is $F_{min}=rac{mu_0 I i pi r}{4R}$ rather than $rac{mu_0 I i pi r^2}{4R}$.

A circular loop of radius

r

carrying a current

i

is held at the centre of another circular loop of radius

R \gg r)

carrying a current

I

. The plane of the smaller loop makes an angle of 30� with that of the larger loop. If the smaller loop is held fixed in this position by applying a single force at a point on its periphery, what would be the minimum magnitude of this force?

Step-by-Step Verified Solution

**Answer:**

F_{min} = \frac{\mu_0 I i \pi r^2}{4R}

. **Step-by-Step Explanation:** Torque on smaller loop:

\tau = \mu B \sin 30� = \frac{\mu_0 I i \pi r^2}{2R} \times \frac{1}{2} = \frac{\mu_0 I i \pi r^2}{4R}

. For equilibrium, force at periphery:

F = \tau/r = \frac{\mu_0 I i \pi r}{4R}

.

All Questions in this Chapter

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