Verified Solution • 2025–26 Session moderate

QPages 230-234: Q8: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q8 Solution: A current of 2 A enters at the corner of a square frame of side …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume the full 2 A current flows through each individual side, failing to recognize that the current splits equally (1 A per branch) at the entry node due to the symmetric resistance of the square frame's two parallel paths.

A current of 2 A enters at the corner dd of a square frame abcdabcd of side 20 cm and leaves at the opposite corner bb. A magnetic field B=0.1 TB = 0.1 \text{ T} exists in the space in a direction perpendicular to the plane of the frame, as shown in the figure. Find the magnitude and direction of the magnetic forces on the four sides of the frame.

Step-by-Step Verified Solution

**Answer:** Forces on sides: Fab=Fcd=0F_{ab} = F_{cd} = 0, Fbc=Fda=0.2 NF_{bc} = F_{da} = 0.2 \text{ N}. **Step-by-Step Explanation:** The current enters at d and leaves at b. By symmetry, the currents in sides ad and dc are equal, as are those in ab and bc. The magnetic field is perpendicular to the plane. Forces on sides parallel to the field are zero. Forces on sides perpendicular to the field are F=IlBF = I l B. For sides dc and ab (length 20 cm), I=1 AI = 1 \text{ A}: F=1×0.2×0.1=0.02 NF = 1 \times 0.2 \times 0.1 = 0.02 \text{ N}.

All Questions in this Chapter

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