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A current of 4×10-3A is flowing in a long straight conductor.the value of the line integral of magnetic field around the closed path enclosing the straight conductor will be. A)1.6π × 10-9 wbm-2 B)1.6 × 10-9 wbm-2 C)1.6 × 10-8 wbm-2 D)1.6π ×10-7 wbm-2?
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A current of 4×10-3A is flowing in a long straight conductor.the value...
Calculation of Line Integral of Magnetic Field
Given:
Current flowing in a long straight conductor = 4×10-3A
To find: Value of line integral of magnetic field around the closed path enclosing the straight conductor
Solution:
- Ampere’s Law states that the line integral of magnetic field around a closed path is directly proportional to the current enclosed by the path.
- Mathematically, the equation for Ampere’s Law is:
∮B.dl = μ0I
where B is the magnetic field, dl is the line element along the path, μ0 is the permeability of free space, and I is the current enclosed by the closed path.
- In this case, the closed path encloses the straight conductor. Therefore, the current enclosed by the path is equal to the current flowing in the conductor, which is 4×10-3A.
- Therefore, the line integral of magnetic field around the closed path is given by:
∮B.dl = μ0I = 4π×10-7×4×10-3
- On solving, we get:
∮B.dl = 1.6π × 10-9 wbm-2
- Hence, the correct option is A) 1.6π × 10-9 wbm-2.
Note: It is important to remember the equation for Ampere’s Law and the concept of current enclosed by a closed path. Also, make sure to use the correct values of the constants involved in the equation.
Given:
Current flowing in a long straight conductor = 4×10-3A
To find: Value of line integral of magnetic field around the closed path enclosing the straight conductor
Solution:
- Ampere’s Law states that the line integral of magnetic field around a closed path is directly proportional to the current enclosed by the path.
- Mathematically, the equation for Ampere’s Law is:
∮B.dl = μ0I
where B is the magnetic field, dl is the line element along the path, μ0 is the permeability of free space, and I is the current enclosed by the closed path.
- In this case, the closed path encloses the straight conductor. Therefore, the current enclosed by the path is equal to the current flowing in the conductor, which is 4×10-3A.
- Therefore, the line integral of magnetic field around the closed path is given by:
∮B.dl = μ0I = 4π×10-7×4×10-3
- On solving, we get:
∮B.dl = 1.6π × 10-9 wbm-2
- Hence, the correct option is A) 1.6π × 10-9 wbm-2.
Note: It is important to remember the equation for Ampere’s Law and the concept of current enclosed by a closed path. Also, make sure to use the correct values of the constants involved in the equation.
Community Answer
A current of 4×10-3A is flowing in a long straight conductor.the value...
According to ampere's law,
B.dl=4π ×10-7 ×4 ×10-3
=1.6π ×10-9 wbm-2
B.dl=4π ×10-7 ×4 ×10-3
=1.6π ×10-9 wbm-2
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A current of 4×10-3A is flowing in a long straight conductor.the value of the line integral of magnetic field around the closed path enclosing the straight conductor will be. A)1.6π × 10-9 wbm-2 B)1.6 × 10-9 wbm-2 C)1.6 × 10-8 wbm-2 D)1.6π ×10-7 wbm-2?
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A current of 4×10-3A is flowing in a long straight conductor.the value of the line integral of magnetic field around the closed path enclosing the straight conductor will be. A)1.6π × 10-9 wbm-2 B)1.6 × 10-9 wbm-2 C)1.6 × 10-8 wbm-2 D)1.6π ×10-7 wbm-2? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A current of 4×10-3A is flowing in a long straight conductor.the value of the line integral of magnetic field around the closed path enclosing the straight conductor will be. A)1.6π × 10-9 wbm-2 B)1.6 × 10-9 wbm-2 C)1.6 × 10-8 wbm-2 D)1.6π ×10-7 wbm-2? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A current of 4×10-3A is flowing in a long straight conductor.the value of the line integral of magnetic field around the closed path enclosing the straight conductor will be. A)1.6π × 10-9 wbm-2 B)1.6 × 10-9 wbm-2 C)1.6 × 10-8 wbm-2 D)1.6π ×10-7 wbm-2?.
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