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Lenz's law is a consequence of the law of conservation of

Charge

Momentum

Mass

Energy

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Explanation

The energy of the field increases with the magnitude of the field. Lenz’s law infers that there is an opposite field created due to increase or decrease of magnetic flux around a conductor so as to hold the law of conservation of energy.

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The magnetic flux through a circuit of resistance R changes by an amount $Δ φ$ in time $Δ t$ , Then the total quantity of electric charge Q, which passing during this time through any point of the circuit is given by

$Q = \frac{Δ ϕ}{Δ t}$

$Q = \frac{Δ ϕ}{Δ t} \times R$

$Q = - \frac{Δ ϕ}{Δ t} + R$

$Q = \frac{Δ ϕ}{R}$

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Explanation

We know that $e = \frac{d ϕ}{d t}$

But e=iR and $i = \frac{d q}{d t} \Rightarrow \frac{d q}{d t} R = \frac{d ϕ}{d t}$ ⇒ $d q = \frac{d ϕ}{R}$

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A coil having an area A₀ is placed in a magnetic field which changes from B₀to 4B₀ in a time interval t. The e.m.f. induced in the coil will be

$\frac{3 A_{0} B_{0}}{t}$

$\frac{4 A_{0} B_{0}}{t}$

$\frac{3 B_{0}}{A_{0} t}$

$\frac{4 B_{0}}{A_{0} t}$

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Explanation

$e = - \frac{d ϕ}{d t} = \frac{- 3 B_{0} A_{0}}{t}$

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A copper ring is held horizontally and a bar magnet is dropped through the ring with its length along the axis of the ring. The acceleration of the falling magnet while it is passing through the ring is

Equal to that due to gravity

Less than that due to gravity

More than that due to gravity

Depends on the diameter of the ring and the length of the magnet

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A magnet is brought towards a coil (i) speedily (ii) slowly then the induced e.m.f./induced charge will be respectively

More in first case / More in the first case

More in first case/Equal in both case

Less in first case/More in second case

Less in first case/Equal in both case

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Explanation

The magnitude of induced e.m.f. is directly proportional to the rate of change of magnetic flux. Induced charge doesn’t depend upon time.

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A coil having 500 square loops each of side 10 cm is placed normal to a magnetic flux which increases at the rate of 1.0 tesla/second. The induced e.m.f. in volts is

0.1

0.5

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Explanation

$| e | = N (\frac{Δ B}{Δ t}) . A \cos θ$

$= 500 \times 1 \times {(10 \times 10^{- 2})}^{2} \cos 0 = 5 V .$

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The magnetic field in a coil of 100 turns and 40 square cm area is increased from 1 Tesla to 6 Tesla in 2 second. The magnetic field is perpendicular to the coil. The e.m.f. generated in it is

10⁴ V

1.2 V

1.0 V

10^–2 V

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Explanation

$e = - N (\frac{Δ B}{Δ t}) . A \cos θ$

$= - 100 \times \frac{(6 - 1)}{2} \times (40 \times 10^{- 4}) \cos 0 \Rightarrow | e | = 1 V$

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The total charge induced in a conducting loop when it is moved in the magnetic field depends on

The rate of change of magnetic flux

Initial magnetic flux only

The total change in magnetic flux

Final magnetic flux only

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Explanation

$q = \frac{N}{R} d ϕ$

$∴ q \propto d ϕ$

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A coil having n turns and resistance RΩ is connected with a galvanometer of resistance 4RΩ. This combination is moved in time t seconds from a magnetic field W₁ $w e b e r / m^{2}$ to W₂ $w e b e r / m^{2}$ . The induced current in the circuit is

$- \frac{W_{2} - W_{1}}{5 R n t}$

$- \frac{n (W_{2} - W_{1})}{5 R t}$

$- \frac{(W_{2} - W_{1})}{R n t}$

$- \frac{n (W_{2} - W_{1})}{R t}$

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Explanation

$i = \frac{e}{R} = \frac{- N}{R} \frac{(ϕ_{2} - ϕ_{1})}{Δ t} = \frac{- n (W_{2} - W_{1})}{5 R t}$

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Two rails of a railway track insulated from each other and the ground are connected to a milli voltmeter. What is the reading of voltmeter, when a train travels with a speed of 180 km/hr along the track. Given that the vertical component of earth's magnetic field is 0.2 × 10^–4 weber/m² and the rails are separated by 1 metre

10^–2 volt

10^–4 volt

10^–3 volt

1 volt

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Explanation

$e = B_{v} . v l$ $= 0 .2 \times 10^{- 4} \times (\frac{180 \times 1000}{3600}) \times 1$

$= 10^{- 3} V$

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