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4 Questions around this concept.
A toroidal solenoid has 3000 turns and a mean radius of 10 cm. It has a soft iron core of relative permeability 2000. What is the magnitude of the magnetic field in the core when a current of 1 A is passed through the solenoid?
A loosely wound helix made of stiff wire is mounted vertically with the lower end just touching a dish of mercury. When a current from the battery is started in the coil through the mercury :
A toroid of mean radius ' a ', cross-section radius ' r ' and a total number of turns N. it carries a current ' i '. The torque experienced by the toroid if a uniform magnetic field of strength B is applied.
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Toroid -
If we try to bend a solenoid in the form of a ring then the obtained shape is a Toroid. So, a toroid can be considered a ring-shaped closed solenoid. Hence it is like an endless cylindrical solenoid. From the given figure we can understand Toroid much better.
Now to obtain the magnetic field by a toroid, let us consider a toroid having N turns.
Here, we will now apply Ampere circuital law to calculate the magnetic field of a toroid. Suppose we have to find the magnetic field B at a point P inside the toroid as shown below in figure -
Let us take an amperian loop which is a circle through point P and concentric inside the toroid. By symmetry, field will have equal magnitude at all points of this circle and this field is tangential to every point in the circle
Thus,
From the above result, B varies with r i.e. field B is not uniform over the cross-section of the core because as we increase 'r' the B varies.
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