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QPages 230-234: Q21: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q21 Solution: Figure shows a rod PQ of length 20.0 cm and mass 200 g…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently misapply the right-hand rule or assume the magnetic force always adds to the weight, leading to an incorrect increase in tension. You must determine the force direction via the cross product $\vec{F} = I\vec{L} \times \vec{B}$; if the force is upward, it opposes gravity, so the tension is calculated as $T = (mg - F_B)/2$.

Figure shows a rod PQ of length 20.0 cm and mass 200 g suspended through a fixed point O by two threads of lengths 20.0 cm each. A magnetic field of strength 0.500 T exists in the vicinity of the wire PQ, as shown in the figure. The wires connecting PQ with the battery are loose and exert no force on PQ. a) Find the tension in the threads when the switch S is open. b) A current of 2.0 A is established when the switch S is closed. Find the tension in the threads now.

Step-by-Step Verified Solution

**Answer:** (a)

T = 1 \text{ N}

(b)

T = 0.8 \text{ N}

. **Step-by-Step Explanation:** (a) Switch open:

T = mg/2 = 0.2 \times 10/2 = 1 \text{ N}

. (b) Switch closed: Magnetic force

F = i l B = 2 \times 0.2 \times 0.5 = 0.2 \text{ N}

upward. New tension

T' = (mg - F)/2 = (2 - 0.2)/2 = 0.9 \text{ N}

.

All Questions in this Chapter

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