English

Obtain the Expression for Mutual Inductance of a Pair of Long Coaxial Solenoids - Physics

Advertisements
Advertisements

Question

Obtain the expression for mutual inductance of a pair of long coaxial solenoids each of length l and radii r1 and r2 (r2 >> r1). Total number of turns in the two solenoids are N1 and N2, respectively.

Advertisements

Solution

Suppose a current i is passed through the inner solenoid S1.

A magnetic field B=μ0n1i is produced inside S1 , whereas the field outside it is zero.

The flux through each turn S2 is Bπr12=μ0n1iπr12

The total flux through all the turns in a length l of S2 is

ϕ=(μ0n1iπr12)n2l=(μ0n1n2πr12l)i

M=μ0n1n2πr12l

shaalaa.com
  Is there an error in this question or solution?
2016-2017 (March) Delhi Set 1

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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
on the axial line, distance ‘r’ from the centre of the solenoid. How does this expression compare with the axial magnetic field due to a bar magnet of magnetic moment ‘m’?


An observer to the left of a solenoid of N turns each of cross section area 'A' observes that a steady current I in it flows in the clockwise direction. Depict the magnetic field lines due to the solenoid specifying its polarity and show that it acts as a bar magnet of magnetic moment m = NIA.

 


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 12 A and is lying on the table. Another wire CD carrying 5 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]


How is the magnetic field inside a given solenoid made strong?


The magnetic field inside a tightly wound, long solenoid is B = µ0 ni. It suggests that the field does not depend on the total length of the solenoid, and hence if we add more loops at the ends of a solenoid the field should not increase. Explain qualitatively why the extra-added loops do not have a considerable effect on the field inside the solenoid.  


A circular coil of one turn of radius 5.0 cm is rotated about a diameter with a constant angular speed of 80 revolutions per minute. A uniform magnetic field B = 0.010 T exists in a direction perpendicular to the axis of rotation. Suppose the ends of the coil are connected to a resistance of 100 Ω. Neglecting the resistance of the coil, find the heat produced in the circuit in one minute.


The magnetic field B inside a long solenoid, carrying a current of 5.00 A, is 3.14 × 10−2 T. Find the number of turns per unit length of the solenoid. 


A capacitor of capacitance 100 µF is connected to a battery of 20 volts for a long time and then disconnected from it. It is now connected across a long solenoid having 4000 turns per metre. It is found that the potential difference across the capacitor drops to 90% of its maximum value in 2.0 seconds. Estimate the average magnetic field produced at the centre of the solenoid during this period. 


A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of magnetic field will be equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×