A wire of fixed length is wound on a solenoid of length ‘l ’ and radius ‘r ’. Its self inductance is found to be L. Now if same wire is wound on a solenoid of length l/2 and radius r/2 then the self inductance will be nL. Find the value of n.
Length of wire = N2πr = Constant (= C, suppose)
∴ Self inductance will become 2L.
The correct answer is: 2
The number of turns, cross-sectional area and length for four solenoids are given in the following table.
The solenoid with maximum self inductance is :
By putting the given values, it can be seen that it is maximum for solenoid number 4.
The correct answer is: 4
In an ideal transformer, the voltage and the current in the primary are 200 volt and 2amp respectively. If the voltage in the secondary is 2000 volt. Then value of current (in Amp) in the secondary will be :
Voltage in primary vp = 200volt
Current in primary ip = 2amp
Voltage in secondary vs = 200volt
The relation for the current in the secondary is
The correct answer is: 0.2
In a L-R growth circuit, inductance and resistance used are 1H and 20Ω respectively. If at t = 50 millisecond, current in the circuit is 3.165A then applied direct current emf (in volt) is :
The correct answer is: 100
A long coaxial cable consists of two thin walled conducting cylinders with inner radius 2cm and outer radius 8cm. The inner cylinder carries a steady current 1A, and the outer cylinder provides the return path for that current. The current produces a magnetic field between the two cylinders. Find the energy stored in the magnetic field for length 1m of the cable. Express answer in nJ (use ln2 = 0.7)
The magnetic field inside is only due to current of inner cylinder.
Magnetic field energy density is not uniform in the space in between cylinders. At distance r from the centre
Energy in volume of element (length l)
Using values U = 140 nJ
The correct answer is: 140
A uniform magnetic field exists in region given by A rod of length 5m is placed along y-axis is moved along x-axis with constant speed 1m/sec. Then induced emf (in volts) in the rod will be :
The correct answer is: 25
Figure shows a square loop of side 1m and resistance 1Ω. The magnetic field on left side of line PQ has a magnitude B = 1.0T. The work done (in Joule) in pulling the loop out of the field uniformly in 1s is
The correct answer is: 1
The magnetic field of a cylindrical magnet that has a pole-face radius 2.8cm can be varied sinusoidally between minimum value 16.8T and maximum value 17.2T at a frequency of Cross section of the magnetic field created by the magnet is shown. At a radial distance of 2cm from the axis find the amplitude of the electric field (in mN/C) induced by the magnetic field variation.
Magnitude of the amplitude
The correct answer is: 240
In the figure, a long thin wire carrying a varying current lies at a distance y above one edge of a rectangular wire loop of length L and width W lying in the x-z plane. If emf is induced in the loop. Find the value of (A+B+C)
The magnetic field at point P, is
The magnetic flux through the shaded strip, is
∴ total magnetic flux through rectangular loop is
∴ induced emf in the loop is
The correct answer is: 8
In the figure shown ABCDEFGH is a square conducting frame of side 2m and resistance 1Ω/m·A uniform magnetic field B is applied perpendicular to the plane and pointing inwards. It increases with time at a constant rate of 10T/s. Find the rate at which heat in watt is produced in the circuit AB = BC = CD = BH.
The induced emf in loop ABHFG
The induced emf in loop BCDH and DEFH
= 1 × 10 = 10 volt KVL in top left loop is
= 1 × 10 = 10 volt
10 – (y – z) + (x – y) - 2y = 0
⇒ x – 4y + z = – 10 ...(1)
KVL in right loop : 10 – 2x – (x – y) – (x – z) = 0
⇒ –4x + y + z = –10 ...(2)
By equation (1) and (2) it is seen that x – y = 0 ⇒ no current in DH
(This can also be seen by symmetry)
This makes solution very simple now the circuit is,
Assume VB = 0, and VF = V, then
⇒ V = 0 ⇒ no current in FB
∴ circuit is :
∴ rate of heat production = (40)2/8 = 200 Watt.
Note : If you are not able to observe the symmetry or decide
x – y = 0, then write KVL in the lower loop. It will be x + y – 6z = –20 ...(3)
Solving (1), (2) and (3) you will get x = +5, y = 5, z = 5A. Heat rate will be
The correct answer is: 40