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प्रश्न
Define the term self-inductance of a solenoid.
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उत्तर
The ratio of magnetic flux through the solenoid to the current passing through it is called self-inductance of a solenoid. It is given by
`L=phi/i`
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संबंधित प्रश्न
Derive the expression for the magnetic field due to a solenoid of length ‘2l’, radius ‘a’ having ’n’ number of turns per unit length and carrying a steady current ‘I’ at a point
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Derive an expression for the mutual inductance of two long co-axial solenoids of same length wound one over the other,
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A magnetic field of 100 G (1 G = 10−4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10−3 m2. The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m−1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.
A wire AB is carrying a steady current of 6 A and is lying on the table. Another wire CD carrying 4 A is held directly above AB at a height of 1 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB. [Take the value of g = 10 ms−2]
Define mutual inductance between two long coaxial solenoids. Find out the expression for the mutual inductance of inner solenoid of length l having the radius r1 and the number of turns n1 per unit length due to the second outer solenoid of same length and r2 number of turns per unit length.
A long solenoid of radius 2 cm has 100 turns/cm and carries a current of 5 A. A coil of radius 1 cm having 100 turns and a total resistance of 20 Ω is placed inside the solenoid coaxially. The coil is connected to a galvanometer. If the current in the solenoid is reversed in direction, find the charge flown through the galvanometer.
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