Magnetic field due to currentMedium
Question
A solenoid is 1.5 m long its inner diameter is 4.0 cm It has 3 layers of windings of 1000 turns each and carries a current of 2.0 amperes. The magnetic flux for a cross-section of the solenoid is nearly
Options
A.4.1 × 10-5 weber
B.5.2 × 10-5 weber
C.6.31 × 10-3 weber
D.2.5 × 10-7 weber
Solution
Magnetic flux (φ) = nBA
where n is number of turns, B is magnetic field and A is area
Given, n = 1000, i = 2 amp r = 0.02 and l = 1.5 m
Magnetic field of solenoid of length l is
B =
∴ φ =
φ =
= 6.31 × 10-3 Wb
where n is number of turns, B is magnetic field and A is area
Given, n = 1000, i = 2 amp r = 0.02 and l = 1.5 m
Magnetic field of solenoid of length l is
B =
∴ φ =
φ =
= 6.31 × 10-3 Wb
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