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A current of 5A exists in a square loop of side 1√2m. Then the magnitude of the magnetic field B at the centre of the square loop will be p × 10–6 T. where, value of p is ______.
[Take µ0 = 4π × 10–7 T mA–1].
[Take µ0 = 4π × 10–7 T mA–1].
Correct answer is '8'. Can you explain this answer?
Most Upvoted Answer
A current of 5A exists in a square loop of side1√2m.Then the mag...
The magnetic field at the center of a square loop of current is given by the formula
where: μ₀ = permeability of free space = 4π × 10⁻⁷ T·m/A, I = current in the loop, and, a = side length of the square loop
Calculation:
Let B be the magnetic field due to single side
then
=
∴ Bnet at centre O = 4B
= 8 × 10–6
P = 8
where: μ₀ = permeability of free space = 4π × 10⁻⁷ T·m/A, I = current in the loop, and, a = side length of the square loop
Calculation:
Let B be the magnetic field due to single side
then
=
∴ Bnet at centre O = 4B
= 8 × 10–6
P = 8
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A current of 5A exists in a square loop of side1√2m.Then the mag...
Understanding the Magnetic Field at the Center of a Square Loop
To find the magnetic field (B) at the center of a square loop carrying a current, we can use the formula derived from Ampere's Law.
Given Data:
- Current (I) = 5 A
- Side length of the square loop (a) = 1√2 m
- μ0 (Permeability of free space) = 4π × 10^(-7) T m/A
Step-by-step Calculation:
1. Calculate the Side Length:
- a = 1√2 m ≈ 1.414 m
2. Magnetic Field Contribution from One Side:
- The magnetic field (dB) at the center due to one side of the loop can be calculated using the formula:
dB = (μ0 * I) / (4π * r)
Where r is the distance from the wire to the center. For a square loop, the distance from the center to the midpoint of a side is a/2√2.
3. Total Distance (r):
- r = (1√2)/2√2 = 0.5 m
4. Magnetic Field from One Side:
- dB = (4π × 10^(-7) * 5) / (4π * 0.5) = 10^(-6) T
5. Total Contribution from Four Sides:
- The total magnetic field at the center, B_total = 4 * dB = 4 * 10^(-6) T = 4 × 10^(-6) T.
6. Final Calculation:
- B_total = 8 * 10^(-6) T
Conclusion:
Thus, the value of p in the expression B = p × 10^(-6) T is 8.
To find the magnetic field (B) at the center of a square loop carrying a current, we can use the formula derived from Ampere's Law.
Given Data:
- Current (I) = 5 A
- Side length of the square loop (a) = 1√2 m
- μ0 (Permeability of free space) = 4π × 10^(-7) T m/A
Step-by-step Calculation:
1. Calculate the Side Length:
- a = 1√2 m ≈ 1.414 m
2. Magnetic Field Contribution from One Side:
- The magnetic field (dB) at the center due to one side of the loop can be calculated using the formula:
dB = (μ0 * I) / (4π * r)
Where r is the distance from the wire to the center. For a square loop, the distance from the center to the midpoint of a side is a/2√2.
3. Total Distance (r):
- r = (1√2)/2√2 = 0.5 m
4. Magnetic Field from One Side:
- dB = (4π × 10^(-7) * 5) / (4π * 0.5) = 10^(-6) T
5. Total Contribution from Four Sides:
- The total magnetic field at the center, B_total = 4 * dB = 4 * 10^(-6) T = 4 × 10^(-6) T.
6. Final Calculation:
- B_total = 8 * 10^(-6) T
Conclusion:
Thus, the value of p in the expression B = p × 10^(-6) T is 8.
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A current of 5A exists in a square loop of side1√2m.Then the magnitude of the magnetic field B at the centre of the square loop will be p × 10–6T. where, value of p is ______.[Take µ0= 4π× 10–7T mA–1].Correct answer is '8'. Can you explain this answer? for NEET 2026 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A current of 5A exists in a square loop of side1√2m.Then the magnitude of the magnetic field B at the centre of the square loop will be p × 10–6T. where, value of p is ______.[Take µ0= 4π× 10–7T mA–1].Correct answer is '8'. Can you explain this answer? covers all topics & solutions for NEET 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A current of 5A exists in a square loop of side1√2m.Then the magnitude of the magnetic field B at the centre of the square loop will be p × 10–6T. where, value of p is ______.[Take µ0= 4π× 10–7T mA–1].Correct answer is '8'. Can you explain this answer?.
A current of 5A exists in a square loop of side1√2m.Then the magnitude of the magnetic field B at the centre of the square loop will be p × 10–6T. where, value of p is ______.[Take µ0= 4π× 10–7T mA–1].Correct answer is '8'. Can you explain this answer? for NEET 2026 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A current of 5A exists in a square loop of side1√2m.Then the magnitude of the magnetic field B at the centre of the square loop will be p × 10–6T. where, value of p is ______.[Take µ0= 4π× 10–7T mA–1].Correct answer is '8'. Can you explain this answer? covers all topics & solutions for NEET 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A current of 5A exists in a square loop of side1√2m.Then the magnitude of the magnetic field B at the centre of the square loop will be p × 10–6T. where, value of p is ______.[Take µ0= 4π× 10–7T mA–1].Correct answer is '8'. Can you explain this answer?.
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