What happens to the magnetic field at the center of a circular coil when the number of turns is increased to n, assuming the current remains the same?
The field becomes n times larger
The magnetic field at the centre of a circular loop of radius \(r\) carrying current \(I\) is \(B = \frac{\mu_0 I}{2r}\). A coil of \(n\) turns behaves like \(n\) such loops placed together.
The fields produced by all the turns at the centre are in the same direction, so they add up, giving \(B = \frac{\mu_0 n I}{2r}\). With the current and radius unchanged, the field is directly proportional to the number of turns, so increasing the turns from one to \(n\) makes the field \(n\) times larger.
The field does not stay the same or become weaker, because every extra turn adds its own field in the same direction. It does not grow as \(n^2\) either, because \(B\) depends on the first power of \(n\), not on its square.
Hence, The field becomes n times larger.
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Important Questions from Magnetism
- The magnetic field around a straight current-carrying conductor:
Magnetic field is a quantity that has___________________ .
- The right-hand thumb rule is used to find which of the following?
- How are the magnetic field lines arranged inside a current-carrying solenoid?
- What does it mean when magnetic field lines are drawn closer together in a diagram?
