500 mA की धारा प्रवाहित करने वाले 50 cm लंबे तार को 5 cm चौड़ाई की एक आयताकार कुंडली बनाने के लिए लपेटा जाता है। जब आयताकार कुंडली को 1.5 T के एकसमान चुंबकीय क्षेत्र के लंबवत इसकी लंबी भुजा पर रखा जाता है, तो अनुभव किया जाने वाला बलाघूर्ण ______ होता है।

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  1. 18.75 × 10-3 N m
  2. 33.75 × 10-3 N m
  3. 7.50 × 10-3 N m
  4. 3.75 × 10-3 N m

Answer (Detailed Solution Below)

Option 3 : 7.50 × 10-3 N m
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Detailed Solution

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दिया गया है:

विद्युत धारा, \(I=500mA=500\times10^{-3}A=5\times10^{-1}A\)

चुंबकीय क्षेत्र, \(B=1.5T\)

तार की लंबाई, L \(=50cm\)

आयताकार कुंडली की चौड़ाई \(=5cm=5\times10^{-2}m\)

The area calculated in the solution is correct. Since it is given the length of the wire. The wire is molded into a rectangle. This means the length of the wire will make perimeter of the rectangle. The breadth is given as 5. So, the length could be calculated as 2(l +b) = 50, which implies length as 20cm.

आयताकार कुंडली की लंबाई \(=20cm=20\times10^{-2}m\)

अवधारणा:

किसी धारावाही चालक का बलाघूर्ण निम्न प्रकार दिया जाता है,

  • \(\tau=NIABsin\theta\)

 

स्पष्टीकरण:

\(\theta=90^0\) और N=1

  • \(\tau=NIABsin\theta\) \(=NIABsin90^0=NIAB\)
  • \(A\text { of a rectangular coil }=L\times B= 20\times5\text{ cm}^2=100\times 10^{-4} m^2\)

 

दिए गए सभी मानों को बलाघूर्ण के उपरोक्त समीकरण में रखने पर, हम प्राप्त करते हैं,

\(\tau=5\times10^{-1}\times100\times10^{-4}\times1.5=7.5\times10^{-3}Nm\)

अतः सही उत्तर \(\tau=7.5\times10^{-3}Nm\) है।

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