Advertisements
Advertisements
प्रश्न
Two long coaxial insulated solenoids, S1 and S2 of equal lengths are wound one over the other as shown in the figure. A steady current "I" flow thought the inner solenoid S1 to the other end B, which is connected to the outer solenoid S2 through which the same current "I" flows in the opposite direction so as to come out at end A. If n1 and n2 are the number of turns per unit length, find the magnitude and direction of the net magnetic field at a point (i) inside on the axis and (ii) outside the combined system
Advertisements
उत्तर
(i) The magnetic field due to a current carrying solenoid:
B=μ0ni
where, n = number of turns per unit length
i = current through the solenoid
Now, the magnetic field due to solenoid S1 will be in the upward direction and the magnetic field due to S2 will be in the downward direction (by right-hand screw rule).
Bnet=BS1−BS2
⇒Bnet=μ0n1I−μ0n2I
=μ0I(n1−n2)
In the upward direction
(ii) The magnetic field is zero outside a solenoid.
APPEARS IN
संबंधित प्रश्न
Define the term self-inductance of a solenoid.
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.
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]
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.
A tightly-wound, long solenoid carries a current of 2.00 A. An electron is found to execute a uniform circular motion inside the solenoid with a frequency of 1.00 × 108 rev s−1. Find the number of turns per metre in the solenoid.
The length of a solenoid is 0.4 m and the number turns in it is 500. A current of 3 amp, is flowing in it. In a small coil of radius 0.01 m and number of turns 10, a current of 0.4 amp. is flowing. The torque necessary to keep the axis of this coil perpendicular to the axis of solenoid will be ______.
