The armature of a dc motor has 20W resistance.It draws a current of 1.5 A when run by a 220 V dc supply. The value of the back emf induced in it is

150 V
170 V
180 V
190 V

A rectangular coil of single turn, having area A, rotates in a uniform magnetic field B with an angular velocity ω about an axis perpendicular to the field. If initially the plane of the coil is perpendicular to the field, then the average induced emf when it has rotated through 90° is

\(\frac{\omega BA}{\pi }\)
\(\frac{\omega BA}{2\pi }\)
\(\frac{\omega BA}{4\pi }\)
\(\frac{2\omega BA}{\pi }\)

Solution

Initially flux, φ = BA cos θ = BA

After rotating through an angle 90°.

Flux through the coil is zero.

So, Δφ = BA

The mutual inductance of a pair of coils, each of N turns, is M henry. If a current of I ampere in one of the coils is brought to zero in t second, the emf induced per turn in the other coil, in volt,will be

\(\frac{MI}{t}\)
\(\frac{NMI}{t}\)
\(\frac{MI}{It}\)
\(\frac{MI}{Nt}\)

Solution

A coil has 200 turns and area of 70 cm². The magnetic field perpendicular to the plane of the coil is 0.3 Wb/m² and take 0.1 sec to rotate through 180º.The value of the induced e.m.f. will be

8.4 V
84 V
42 V
4.2 V

Solution

A copper disc of radius 0.1 m rotated about its centre with 10 revolutions per second in a uniform magnetic field of 0.1 tesla with its plane perpendicular to the field. The e.m.f.induced across the radius of disc is

^π⁄₁₀ volt
^2π⁄₁₀ volt
π × 10^-2 volt
2π × 10^-2 volt

Solution

When the current in a coil changes from 2 amp.to 4 amp. in 0.05 sec., an e.m.f. of 8 volt is induced in the coil. The coefficient of self inductance of the coil is

0.1 henry
0.2 henry
0.4 henry
0.8 henry

Solution

A generator has an e.m.f.of 440 Volt and internal resistance of 4000 hm. Its terminals are connected to a load of 4000ohm. The voltage across the load is

220 volt
440 volt
200 volt
400 volt

Solution

Total resistance of the circuit = 4000 + 400= 4400 W

Current flowing i = \(\frac{V}{R}=\frac{440}{4400}\) = 0.1 amp.

Voltage across load = R i= 4000 × 0.1 =400 volt.

In a coil of area 10 cm² and 10 turns with magnetic field directed perpendicular to the plane and is changing at the rate of 10⁸ Gauss/second. The resistance of the coil is 20W.The current in the coil will be

0.5 A
5 A
50 A
5 × 10⁸ A

Solution

Fig shown below represents an area A =0.5 m² situated in a uniform magnetic field B = 2.0 weber/m² and making an angle of 60º with respect to magnetic field.

The value of the magnetic flux through the area would be equal to

2.0 weber
√3 weber
√3 / 2 weber
0.5 weber

Solution

φ = BA cos θ = 2.0 × 0.5 × cos 60° = \(\frac{2.0\times 0.5}{2}=0.5\) weber

A coil having an area A₀ is placed in a magnetic field which changes from B₀ to 4 B₀ in time interval t. The e.m.f. induced in the coil will be

3 A₀ B₀ / t
4 A₀ B₀ / t
3 B₀ / A₀ t
4 A₀ / B₀ t

Solution

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