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QPages 230-234: Q17: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q17 Solution: A semicircular wire of radius 5.0 cm carries a current of 5.0…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat a curved segment as a straight wire of length equal to the diameter and apply $F=ILB$. For a semicircle in a uniform field perpendicular to its plane, the magnetic forces on opposite elements are equal and opposite, so the vector sum is zero. The correct approach is to integrate $dmathbf{F}=I,dmathbf{l} imesmathbf{B}$ and note the symmetry that yields $mathbf{F}_{ ext{net}}=0$.

A semicircular wire of radius 5.0 cm carries a current of 5.0 A. A magnetic field

B

of magnitude 0.50 T exists along the perpendicular to the plane of the wire. Find the magnitude of the magnetic force acting on the wire.

Step-by-Step Verified Solution

**Answer:**

F = 0.125 \text{ N}

. **Step-by-Step Explanation:** For a semicircular wire, the effective length is the diameter

2R

.

F = i (2R) B = 5 \times (2 \times 0.05) \times 0.50 = 0.25 \text{ N}

. Wait, the force on a semicircular wire is

F = 2 i R B = 2 \times 5 \times 0.05 \times 0.50 = 0.25 \text{ N}

.

All Questions in this Chapter

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