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A current carrying wire (current =i=i ) perpendicular to the plane of the paper produces a magnetic field, as shown in the figure. A square of side aa is drawn with one of its vertices on the centre of the wire. The integral ∫→Bd.r along OPQRO
has the value?
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A current carrying wire (current =i=i ) perpendicular to the plane of ...
Introduction:
In this question, we are given a current-carrying wire perpendicular to the plane of the paper. The wire produces a magnetic field, and we need to find the value of the integral of the magnetic field along a square with one of its vertices at the center of the wire.

Given:
- Current in the wire: i
- Side of the square: a

Approach:
To find the integral of the magnetic field along the square, we can divide the square into smaller segments and calculate the contribution of each segment to the total integral. We will use the Biot-Savart Law to calculate the magnetic field at each segment.

Calculation:
1. Dividing the square:
- Let the center of the square be O.
- Divide the square into four segments: OP, PQ, QR, and RO.

2. Magnetic field at each segment:
- The magnetic field at each segment can be calculated using the Biot-Savart Law.
- According to the Biot-Savart Law, the magnetic field at a point due to a current-carrying wire is directly proportional to the current and inversely proportional to the distance from the wire.
- Since the wire is perpendicular to the plane of the paper, the magnetic field will be in the form of concentric circles around the wire.
- The magnitude of the magnetic field at a distance r from the wire is given by the formula: B = (μ0 * i) / (2πr), where μ0 is the permeability of free space.

3. Calculation of the integral:
- The integral of the magnetic field along the square can be calculated as the sum of the contributions from each segment.
- Since the magnetic field is perpendicular to the direction of integration along each segment, the dot product of the magnetic field and the infinitesimal displacement vector dr will be zero.
- Therefore, the integral of the magnetic field along each segment will be zero.
- Hence, the value of the integral ∫→Bd.r along OPQRO will be zero.

Conclusion:
The value of the integral ∫→Bd.r along OPQRO is zero. This is because the magnetic field along each segment of the square is perpendicular to the direction of integration, resulting in a zero contribution to the integral.
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A current carrying wire (current =i=i ) perpendicular to the plane of the paper produces a magnetic field, as shown in the figure. A square of side aa is drawn with one of its vertices on the centre of the wire. The integral ∫→Bd.r along OPQRO has the value?
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A current carrying wire (current =i=i ) perpendicular to the plane of the paper produces a magnetic field, as shown in the figure. A square of side aa is drawn with one of its vertices on the centre of the wire. The integral ∫→Bd.r along OPQRO has the value? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A current carrying wire (current =i=i ) perpendicular to the plane of the paper produces a magnetic field, as shown in the figure. A square of side aa is drawn with one of its vertices on the centre of the wire. The integral ∫→Bd.r along OPQRO has the value? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A current carrying wire (current =i=i ) perpendicular to the plane of the paper produces a magnetic field, as shown in the figure. A square of side aa is drawn with one of its vertices on the centre of the wire. The integral ∫→Bd.r along OPQRO has the value?.
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