A square metal loop of side 10 cm and resistance 1 Ωis moved with a constant velocity partly inside a uniform magnetic field of 2 Wbm^–2, directed into the paper, as shown in the figure. The loop is connected to a network of five resistors each of value 3Ω. If a steady current of 1 mA flows in the loop, then the speed of the loop is

0.5 cms^–1
1 cms^–1
2 cms^–1
4 cms^–1

Solution

A wire loop is rotated in a uniform magnetic field about an ax is perpendicular to the field. The direction of the current induced in the loop reverses once each

quarter revolution
half revolution
full revolution
two revolutions

Solution

It is because after every 1/2 revolution the current becomes zero and mode of change in flux changes there after (If before the current becomes zero, the mode of flux change was from left to right then after the current becomes zero the mode of flux change becomes right to left).

A small square loop of wire of side l is placed inside a large square loop of side L(L >> l). The loop are coplanar and their centres coincide. The mutual inductance of the system is proportional is

\(\frac{\iota }{L}\)
\(\frac{\iota^{2} }{L}\)
\(\frac{L}{\iota }\)
\(\frac{L^{2}}{\iota }\)

A metal rod moves at a constant velocity in a direction perpendicular to its length. A constant uniform magnetic field exists in space in a direction perpendicular to the rod as well its velocity. Select correct statements (s) from the following.

The entire rod is at the same potential
There is an electric field in the rod(c)The electric potential is highest at the centre
The electric potential is lowest at its centre and increases towards its ends
The electric potential is lowest at its centre and increases towards its ends

Solution

Due to shifting of electrons, one end of the rod becomes positive and the other end negative. This developes a electric field in the rod.

A coil is wound on a frame of rectangular cross-section. If all the linear dimensions of the frame are increased by a factor 2 and the number of turns perunit length of the coil remains the same,self-inductance of the coil increases by a factor of

Solution

Self inductance = μ₀n² AL = μ₀ n² (l × b) × L

n = Total number of turns/length

L = Length of inductorl = Length of rectangular cross section

b = breadth of rectangular cross-section

So, when all linear dimensions are increased by a factor of 2. The new self inductance becomes L' = 8L.

Consider the situation shown. The wire AB is sliding on fixed rails with a constant velocity. If the wire AB is replaced by semi-circular wire, the magnitude of inducede.m.f. will

increase
decrease
remain the same
increase or decrease depending on whether the semi-circle buldges towards the resistance or away from it.

Solution

E.m.f.will remain same because change in area per unit time will be same in both cases.

A thin circular ring of area A is held perpendicular to a uniform magnetic field of induction B. A small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of the circuit is R. When the ring is suddenly squeezed to zero area, the charge flowing through the galvanometer is

\(\frac{BR}{A}\)
\(\frac{AB}{R}\)
ABR
\(\frac{B^{2}A}{R^{2}}\)

Solution

Two identical induction coils each of inductance L are jointed in series are placed very close to each other such that the winding direction of one is exactly opposite to that of the other, what is the net inductance?

L²
2 L
L /2
zero

Solution

When two inductance coil are joined in series, such that the winding of one is exactly opposite to each other the emf produced in the two coils are out of phase such that they cancel out.

A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth’s magnetic field is 0.2 × 10^–4T, then the e.m.f. developed between the two ends of the conductor is

5mV
50 μV
5 μV
50 mV

Solution

Two coils have a mutual inductance 0.005H. The current changes in first coil according to equation I = I₀ sin ωt where I₀= 10A and ω = 100π radian/sec. The max. value of e.m.f. in second coil is

Solution

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