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Question
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.
Solution
Given:
Length of the rod PQ = 20.0 cm
Mass of the rod, M = 200 g
Length of the two threads, l = 20.0 cm
Applied magnetic field, B = 0.500 T
As per the question,
(a) When the circuit is open:
The weight of the rod is balanced by the tension in the rod. So,
2Tcos30° = Mg
`T = (mg)/(2cos30^circ)`
=`(0.2xx9.8)/(2 (sqrt(3)/2)`
=1.13 N
(b) When the circuit is closed and current flowing through the circuit = 2 A:
Then,
2Tcos 30°= Mg + ilB
= 0.200× 9.8 + 2 × 0.2 × 0.5
= 1.95 + 0.2 = 2.16
⇒ `2T = 2.16xx (2)/(sqrt(3)`
⇒ T = 1.245 = 1.25 N
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