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The no of turn in a long Solenoid is 500 . The area of cross sections is 2*10-3 m2. If the magnetic induction, on passing a current of 2A, through it is 5*10-3 T, the magnitude of magnetic flux connected with it in webers will be ?
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The no of turn in a long Solenoid is 500 . The area of cross sections ...
Given data:
Number of turns in the solenoid (N) = 500
Area of cross-section of the solenoid (A) = 2 * 10^-3 m^2
Current passing through the solenoid (I) = 2 A
Magnetic induction (B) = 5 * 10^-3 T
Calculation:
The magnetic flux connected with the solenoid can be calculated using the formula:
Φ = B * A * N
Substituting the given values into the formula:
Φ = (5 * 10^-3 T) * (2 * 10^-3 m^2) * 500
Simplifying the expression:
Φ = 5 * 2 * 10^-3 * 2 * 10^-3 * 500 T * m^2
Φ = 20 * 10^-6 * 1000 T * m^2
Φ = 20 * 10^-3 T * m^2
Φ = 20 * 10^-3 Wb
Therefore, the magnitude of the magnetic flux connected with the solenoid is 20 * 10^-3 Wb.
Explanation:
- A solenoid is a long coil of wire with multiple turns.
- The magnetic flux through a solenoid depends on the number of turns, the area of the cross-section, and the magnetic induction.
- The formula to calculate the magnetic flux is Φ = B * A * N, where Φ is the magnetic flux, B is the magnetic induction, A is the area of cross-section, and N is the number of turns.
- By substituting the given values into the formula and simplifying the expression, we can find the magnitude of the magnetic flux connected with the solenoid.
- In this case, the magnetic flux is calculated to be 20 * 10^-3 Wb.
Summary:
The magnetic flux connected with the solenoid is 20 * 10^-3 Wb. The magnetic flux depends on the number of turns, the area of cross-section, and the magnetic induction. The formula to calculate the magnetic flux is Φ = B * A * N, where Φ is the magnetic flux, B is the magnetic induction, A is the area of cross-section, and N is the number of turns. By substituting the given values into the formula, the magnitude of the magnetic flux connected with the solenoid can be calculated.
Number of turns in the solenoid (N) = 500
Area of cross-section of the solenoid (A) = 2 * 10^-3 m^2
Current passing through the solenoid (I) = 2 A
Magnetic induction (B) = 5 * 10^-3 T
Calculation:
The magnetic flux connected with the solenoid can be calculated using the formula:
Φ = B * A * N
Substituting the given values into the formula:
Φ = (5 * 10^-3 T) * (2 * 10^-3 m^2) * 500
Simplifying the expression:
Φ = 5 * 2 * 10^-3 * 2 * 10^-3 * 500 T * m^2
Φ = 20 * 10^-6 * 1000 T * m^2
Φ = 20 * 10^-3 T * m^2
Φ = 20 * 10^-3 Wb
Therefore, the magnitude of the magnetic flux connected with the solenoid is 20 * 10^-3 Wb.
Explanation:
- A solenoid is a long coil of wire with multiple turns.
- The magnetic flux through a solenoid depends on the number of turns, the area of the cross-section, and the magnetic induction.
- The formula to calculate the magnetic flux is Φ = B * A * N, where Φ is the magnetic flux, B is the magnetic induction, A is the area of cross-section, and N is the number of turns.
- By substituting the given values into the formula and simplifying the expression, we can find the magnitude of the magnetic flux connected with the solenoid.
- In this case, the magnetic flux is calculated to be 20 * 10^-3 Wb.
Summary:
The magnetic flux connected with the solenoid is 20 * 10^-3 Wb. The magnetic flux depends on the number of turns, the area of cross-section, and the magnetic induction. The formula to calculate the magnetic flux is Φ = B * A * N, where Φ is the magnetic flux, B is the magnetic induction, A is the area of cross-section, and N is the number of turns. By substituting the given values into the formula, the magnitude of the magnetic flux connected with the solenoid can be calculated.
Community Answer
The no of turn in a long Solenoid is 500 . The area of cross sections ...
Magnetic flux =BAN
=5×10^-3 × 2×10^-3 × 500
=5×10^-3
=5×10^-3 × 2×10^-3 × 500
=5×10^-3
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The no of turn in a long Solenoid is 500 . The area of cross sections is 2*10-3 m2. If the magnetic induction, on passing a current of 2A, through it is 5*10-3 T, the magnitude of magnetic flux connected with it in webers will be ?
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The no of turn in a long Solenoid is 500 . The area of cross sections is 2*10-3 m2. If the magnetic induction, on passing a current of 2A, through it is 5*10-3 T, the magnitude of magnetic flux connected with it in webers will be ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The no of turn in a long Solenoid is 500 . The area of cross sections is 2*10-3 m2. If the magnetic induction, on passing a current of 2A, through it is 5*10-3 T, the magnitude of magnetic flux connected with it in webers will be ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The no of turn in a long Solenoid is 500 . The area of cross sections is 2*10-3 m2. If the magnetic induction, on passing a current of 2A, through it is 5*10-3 T, the magnitude of magnetic flux connected with it in webers will be ?.
The no of turn in a long Solenoid is 500 . The area of cross sections is 2*10-3 m2. If the magnetic induction, on passing a current of 2A, through it is 5*10-3 T, the magnitude of magnetic flux connected with it in webers will be ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The no of turn in a long Solenoid is 500 . The area of cross sections is 2*10-3 m2. If the magnetic induction, on passing a current of 2A, through it is 5*10-3 T, the magnitude of magnetic flux connected with it in webers will be ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The no of turn in a long Solenoid is 500 . The area of cross sections is 2*10-3 m2. If the magnetic induction, on passing a current of 2A, through it is 5*10-3 T, the magnitude of magnetic flux connected with it in webers will be ?.
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