An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to:
80 H
0.08 H
0.044 H
D.
I = 10 A, V = 80 V
R = V/I = 80/10 = 8Ω and ω = 50 Hz
For AC circuit, we have
A galvanometer having a coil resistance of 100 Ω gives a full-scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full-scale deflection for a current of 10 A, is:
0.01 Ω
2 Ω
0.1 Ω
3 Ω
A.
0.01 Ω
Maximum voltage that can be applied across the galvanometer coil = 100 Ω x 10^{-3} A = 0.1
If R_{s} is the shunt resistance, then
R_{s} x 10 A = 0.1 V
R_{s} = 0.01 Ω
A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orientations as shown in the figures below:
If there is a uniform magnetic field of 0.3 T in the positive z direction , in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium?
(a) and (b) respectively
(a) and (c) respectively
(b) and (d) respectively
(b) and (c) respectively
C.
(b) and (d) respectively
Since, B is uniform only torque acts on a current carrying loop
As,
A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB andCD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30º. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the origin.
Due to the presence of the current I_{1} at the origin
The magnitude of the magnetic field (2) due to the loop ABCD at the origin (O) is
The forces on AB and DC are zero
The forces on AD and BC are zero
The magnitude of the net force on the loop is given by
The magnitude of the net force on the loop is given by
B.
The forces on AD and BC are zero
The forces on AD and BC are zero because magnetic field due to a straight wire on AD and BC is parallel to elementary length of the loop.
Two coaxial solenoids of different radii carry current I in the same direction. Let be the magnetic force on the inner solenoid due to the outer one and be the magnetic force on the outer solenoid due to the inner one. Then:
A.
Consider the two coaxial solenoids. Due to one of the solenoids magnetic field at the centre of the other can be assumed to be constant
Due to symmetry, forces on upper and lower part of the solenoid will be equal and opposite and hence resultant is zero.
Therefore option (a) is correct.
In a coil of resistance 100 Ω, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is
250 Wb
275 Wb
200 Wb
225 Wb
A.
250 Wb
Two identical wires A and B, each of length ‘l’, carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side ‘a’. If B_{A} and B_{B} are the values of the magnetic field at the centres of the circle and square respectively, then the ratio B_{A} /B_{B} is:
C.
Magnetic field in case of circle of radius R, we have
Magnetic field in case of square of side we get
A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB andCD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30º. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the origin.
The magnitude of the magnetic field (2) due to the loop ABCD at the origin (O) is
zero
B.
Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper as shown. The variation of the magnetic field B along the line XX' is given by
B.
The magnetic field in between because of each of the conductors will be in naturally opposite directions.
∴ Net magnetic field
At x = d, B in between = 0
For x < d, B in between =
For x> d, B in between =
Towards x, net magnetic field will add up and direction will be . Towards' x' b net magnetic field will add up and direction will be ()
Hysteresis loops for two magnetic materials A and B are given below:
These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use:
A for electric generators and transformers.
A for electromagnets and B for electric generators
A for transformers and B for electric generators.
B for electromagnets and transformers.
D.
B for electromagnets and transformers.
Area of the hysteresis loop is proportional to the net energy absorbed per unit volume by the material, as it is taken over a complete cycle of magnetisation.
For electromagnets and transformers, energy loss should be low.
i.e thin hysteresis curves.
Also |B|→0 When H = 0 and |H| should be small when B →0.