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A straight wire of mass 200 g and length 1.5 m carries a current of 2A. It is suspended in mid air by a uniform horizontal magnetic field whose magnitude in Tesla is
- a)2 T
- b)0.65 T
- c)1.3 T
- d)0.55 T
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A straight wire of mass 200 g and length 1.5 m carries a current of 2A...
It has been found that Force (F) acting on a current-carrying conductor placed on a magnetic field, in a direction perpendicular to the direction of the magnetic field, is directly proportional to the current, length of the conductor and magnitude of the field F = k I l B. In SI units, constant k = 1 Also, Force acting on a suspended mass = mg ⇒ mg = I l b
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A straight wire of mass 200 g and length 1.5 m carries a current of 2A...
Given information:
- Mass of the wire (m) = 200 g = 0.2 kg
- Length of the wire (L) = 1.5 m
- Current flowing through the wire (I) = 2 A
To find: The magnitude of the magnetic field (B) in Tesla.
Formula used:
The magnetic force on a current-carrying wire in a magnetic field is given by the formula:
F = BIL sinθ
Where:
- F is the force on the wire,
- B is the magnetic field strength,
- I is the current in the wire,
- L is the length of the wire, and
- θ is the angle between the magnetic field and the wire.
Since the wire is in mid-air and suspended, the force acting on it is only due to the magnetic field. The weight of the wire is balanced by the tension in the wire, so there is no net force in the vertical direction. Therefore, the force due to the magnetic field must be equal and opposite to the weight of the wire.
The weight of the wire can be calculated using the formula:
Weight = mass * gravity
Where:
- The mass of the wire (m) is given as 0.2 kg, and
- The acceleration due to gravity (g) is approximately 9.8 m/s².
Calculation:
Weight = 0.2 kg * 9.8 m/s² = 1.96 N
Since the magnetic force is equal and opposite to the weight, we have:
F = 1.96 N
Using the formula F = BIL sinθ, we can rearrange it to solve for B:
B = F / (IL sinθ)
In this case, the wire is suspended in mid-air, so the angle θ between the magnetic field and the wire is 90°. Therefore, sinθ = 1.
B = F / (IL * 1)
B = F / IL
Substituting the given values:
B = 1.96 N / (2 A * 1.5 m)
B = 1.96 N / 3 A·m
B = 0.65 T
Hence, the magnitude of the magnetic field is 0.65 T, which corresponds to option 'B'.
- Mass of the wire (m) = 200 g = 0.2 kg
- Length of the wire (L) = 1.5 m
- Current flowing through the wire (I) = 2 A
To find: The magnitude of the magnetic field (B) in Tesla.
Formula used:
The magnetic force on a current-carrying wire in a magnetic field is given by the formula:
F = BIL sinθ
Where:
- F is the force on the wire,
- B is the magnetic field strength,
- I is the current in the wire,
- L is the length of the wire, and
- θ is the angle between the magnetic field and the wire.
Since the wire is in mid-air and suspended, the force acting on it is only due to the magnetic field. The weight of the wire is balanced by the tension in the wire, so there is no net force in the vertical direction. Therefore, the force due to the magnetic field must be equal and opposite to the weight of the wire.
The weight of the wire can be calculated using the formula:
Weight = mass * gravity
Where:
- The mass of the wire (m) is given as 0.2 kg, and
- The acceleration due to gravity (g) is approximately 9.8 m/s².
Calculation:
Weight = 0.2 kg * 9.8 m/s² = 1.96 N
Since the magnetic force is equal and opposite to the weight, we have:
F = 1.96 N
Using the formula F = BIL sinθ, we can rearrange it to solve for B:
B = F / (IL sinθ)
In this case, the wire is suspended in mid-air, so the angle θ between the magnetic field and the wire is 90°. Therefore, sinθ = 1.
B = F / (IL * 1)
B = F / IL
Substituting the given values:
B = 1.96 N / (2 A * 1.5 m)
B = 1.96 N / 3 A·m
B = 0.65 T
Hence, the magnitude of the magnetic field is 0.65 T, which corresponds to option 'B'.
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A straight wire of mass 200 g and length 1.5 m carries a current of 2A. It is suspended in mid air by a uniform horizontal magnetic field whose magnitude in Tesla isa)2 Tb)0.65 Tc)1.3 Td)0.55 TCorrect answer is option 'B'. Can you explain this answer? for Class 10 2025 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about A straight wire of mass 200 g and length 1.5 m carries a current of 2A. It is suspended in mid air by a uniform horizontal magnetic field whose magnitude in Tesla isa)2 Tb)0.65 Tc)1.3 Td)0.55 TCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 10 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A straight wire of mass 200 g and length 1.5 m carries a current of 2A. It is suspended in mid air by a uniform horizontal magnetic field whose magnitude in Tesla isa)2 Tb)0.65 Tc)1.3 Td)0.55 TCorrect answer is option 'B'. Can you explain this answer?.
A straight wire of mass 200 g and length 1.5 m carries a current of 2A. It is suspended in mid air by a uniform horizontal magnetic field whose magnitude in Tesla isa)2 Tb)0.65 Tc)1.3 Td)0.55 TCorrect answer is option 'B'. Can you explain this answer? for Class 10 2025 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about A straight wire of mass 200 g and length 1.5 m carries a current of 2A. It is suspended in mid air by a uniform horizontal magnetic field whose magnitude in Tesla isa)2 Tb)0.65 Tc)1.3 Td)0.55 TCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 10 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A straight wire of mass 200 g and length 1.5 m carries a current of 2A. It is suspended in mid air by a uniform horizontal magnetic field whose magnitude in Tesla isa)2 Tb)0.65 Tc)1.3 Td)0.55 TCorrect answer is option 'B'. Can you explain this answer?.
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