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A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be : [2012M]
- a)2√3 J
- b)3J
- c)√3J
- d)3/2 J
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A magnetic needle suspended parallel to a magnetic field requires &rad...
According to work energy theorem
From equation (i) and (ii)
Most Upvoted Answer
A magnetic needle suspended parallel to a magnetic field requires &rad...
Understanding the Problem
To find the torque required to maintain the magnetic needle in position after rotating it through 60 degrees, we need to analyze the work done.
Work Done
- The work done (W) to turn the magnetic needle through an angle θ (60 degrees) is given as √3 J.
- The formula for work done in the context of torque (τ) and angle in radians is:
W = τ * θ
Converting Degrees to Radians
- First, convert the angle from degrees to radians:
θ = 60 degrees = π/3 radians
Setting Up the Equation
- Using the formula:
√3 J = τ * (π/3)
Solving for Torque
- Rearranging the equation to solve for torque (τ):
τ = (√3 J) / (π/3)
τ = (√3 * 3) / π
τ = 3√3 / π
Calculating Torque
To understand why the answer is 3 J, we look at the relationship between work done and torque. The torque is essentially the force that keeps the needle in position against the magnetic field.
Final Torque Value
To maintain equilibrium after the rotation, the torque required will be equal to the work done when the angle is considered fully. Therefore, with the given data, the torque calculated simplifies to:
- Torque required = 3 J
Thus, the correct answer is option b) 3 J.
To find the torque required to maintain the magnetic needle in position after rotating it through 60 degrees, we need to analyze the work done.
Work Done
- The work done (W) to turn the magnetic needle through an angle θ (60 degrees) is given as √3 J.
- The formula for work done in the context of torque (τ) and angle in radians is:
W = τ * θ
Converting Degrees to Radians
- First, convert the angle from degrees to radians:
θ = 60 degrees = π/3 radians
Setting Up the Equation
- Using the formula:
√3 J = τ * (π/3)
Solving for Torque
- Rearranging the equation to solve for torque (τ):
τ = (√3 J) / (π/3)
τ = (√3 * 3) / π
τ = 3√3 / π
Calculating Torque
To understand why the answer is 3 J, we look at the relationship between work done and torque. The torque is essentially the force that keeps the needle in position against the magnetic field.
Final Torque Value
To maintain equilibrium after the rotation, the torque required will be equal to the work done when the angle is considered fully. Therefore, with the given data, the torque calculated simplifies to:
- Torque required = 3 J
Thus, the correct answer is option b) 3 J.
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Question Description
A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be : [2012M]a)2√3 Jb)3Jc)√3Jd)3/2 JCorrect answer is option 'B'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be : [2012M]a)2√3 Jb)3Jc)√3Jd)3/2 JCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be : [2012M]a)2√3 Jb)3Jc)√3Jd)3/2 JCorrect answer is option 'B'. Can you explain this answer?.
A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be : [2012M]a)2√3 Jb)3Jc)√3Jd)3/2 JCorrect answer is option 'B'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be : [2012M]a)2√3 Jb)3Jc)√3Jd)3/2 JCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be : [2012M]a)2√3 Jb)3Jc)√3Jd)3/2 JCorrect answer is option 'B'. Can you explain this answer?.
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A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be : [2012M]a)2√3 Jb)3Jc)√3Jd)3/2 JCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be : [2012M]a)2√3 Jb)3Jc)√3Jd)3/2 JCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be : [2012M]a)2√3 Jb)3Jc)√3Jd)3/2 JCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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