The magnetic field due to a current-carrying solenoid which has 'n' number of turns per unit length is:
Correct answer: A. B=μonI
- A. B=μonI
- B. B=μon2I
- C. B= 4μonI
- D. B= 4μon2I
Explanation
The magnetic field is given by the formulaB = μ0nIWhereB is the magnetic fieldn is the number of turns in the solenoidI is the currentμ0 is the permeability of free space = 4π x 10-7
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About Electromagnetism
Magnetic flux density and magnetic flux describe the strength of a magnetic field and the field passing through a surface. A charged particle moving through a magnetic field experiences a force perpendicular to its velocity and may follow circular or helical motion, depending on the angle between velocity and field.
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