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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 - Physics

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प्रश्न

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 is open. (b) A current of 2.0 A is established when the switch S is closed. Find the tension in the threads now.

बेरीज
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उत्तर


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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पाठ 12: Magnetic Field - Exercises [पृष्ठ २३१]

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एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
पाठ 12 Magnetic Field
Exercises | Q 21 | पृष्ठ २३१

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