Question
Question: A toroidal solenoid with an air core has an average radius of 15cm, if area of cross section \[12\,c...
A toroidal solenoid with an air core has an average radius of 15cm, if area of cross section 12cm2 and 1200 turns. Ignoring the field variation across the area of cross section of the toroid, the self inductance of the toroid will be-
(A). 4.6mH
(B). 6.9mH
(C). 2.3mH
(D). 9.2mH
Solution
Inductance of a conductor opposes the change in current of its circuit. A solenoid consists of wire wound around an armature. Its inductance is related to a number of factors like number of turns, area of cross section, radius etc. used in the relationship of all parameters to find the inductance.
Formula used:
L=2πrμ0N2A
Complete step by step solution:
Self induction is the induction of a voltage in a current carrying wire when the current in the circuit is continuously changing. Here, the voltage is induced due to change in the current of its own circuit. Due to changing current the magnetic field associated with the circuit also changes and a potential difference is induced. The potential induced is given by-
e=−LdtdI
e is the potential induced
L is the inductance of the circuit
dtdI is the rate of change of current in the circuit
The negative sign indicates current induced is in a direction opposite to the direction of changing current.
The formula for inductance for a solenoid is given by-
L=2πrμ0N2A
Here, L is the inductance of solenoid
μ0 is the permeability of free air
N is the total number of turns
A is the area of cross section
r is the radius
Substituting given values in the above equation, we get,