12. The magnetic flux through a circuit of resistance $R$ changes by an amount $\Delta \phi$ in a time $\Delta t$. Then the total quantity of electric charge $Q$ that passes any point in the circuit during the time $\Delta t$ is represented by: (2004)
The induced emf is $e = \frac{\Delta \phi}{\Delta t}$.
The induced current is $I = \frac{e}{R} = \frac{\Delta \phi}{R \Delta t}$.
Total charge $Q = I \Delta t = \frac{\Delta \phi}{R \Delta t} \cdot \Delta t = \frac{\Delta \phi}{R}$.
14. Initially plane of coil is parallel to the uniform magnetic field $B$. In time $\Delta t$ it makes to perpendicular to the magnetic field, then charge flows in $\Delta t$ depends on this time as: (1999)
The total charge flowing through the circuit is given by $Q = \frac{\Delta \Phi}{R}$.
This expression shows that the induced charge depends only on the net change in magnetic flux and resistance.
It is independent of the time interval $\Delta t$, meaning it is proportional to $(\Delta t)^0$.
15. The total charge, induced in a conducting loop when it is moved in magnetic field depend on (1992)
Induced charge $Q = \int I dt = \int \frac{e}{R} dt = \int \frac{1}{R} \frac{d\Phi}{dt} dt = \frac{\Delta \Phi}{R}$.
Thus, the total induced charge depends directly on the total change in magnetic flux, $\Delta \Phi$.
It is independent of the rate of change of flux or time.
16. A rectangular coil of $20$ turns and area of cross-section $25 sq.cm$ has a resistance of $100 \Omega$. If a magnetic field which is perpendicular to the plane of coil changes at a rate of $1000 tesla per second$, the current in the coil is (1992)
Induced emf $e = N A \frac{dB}{dt} = 20 \times (25 \times 10^{-4}) \times 1000 = 50 V$.
Induced current $I = \frac{e}{R}$.
$I = \frac{50}{100} = 0.5 A$.
17. Faraday’s laws are consequence of conservation of (1991)
Faraday's laws of electromagnetic induction, and specifically Lenz's law which determines the direction of induced emf, are based on the principle of conservation of energy.
The mechanical work done in moving a magnet or coil against the opposing magnetic force is converted into electrical energy.
18. A magnetic field of $2 \times 10^{-2} T$ acts at right angles to a coil of area $100 cm^2$, with $50$ turns. The average e.m.f. induced in the coil is $0.1 V$, when it is removed from the field in $t sec$. The value of $t$ is (1991)
Initial flux $\Phi_i = N B A = 50 \times (2 \times 10^{-2}) \times (100 \times 10^{-4}) = 10^{-2} Wb$.
Final flux $\Phi_f = 0$. Magnitude of average emf $e = \frac{\Delta \Phi}{t} = \frac{10^{-2}}{t}$.
Given $e = 0.1 V$, we have $0.1 = \frac{10^{-2}}{t}$, which gives $t = \frac{10^{-2}}{0.1} = 0.1 s$.
22. A big circular coil of 100 turns and average radius $10 m$ is rotating about its horizontal diameter at $2 rad s^{-1}$. If the vertical component of earth’s magnetic field at that place is $2 \times 10^{-5} T$ and electrical resistance of the coil is $12.56 \Omega$, then the maximum induced current in the coil will be : (2022)
The peak induced emf is given by $e_0 = N B A \omega = N B (\pi r^2) \omega$.
Substituting the values: $e_0 = 100 \times (2 \times 10^{-5}) \times (\pi \times 10^2) \times 2 = 4\pi \approx 12.56 V$.
The maximum induced current is $I_0 = \frac{e_0}{R} = \frac{12.56}{12.56} = 1 A$.
23. A wheel with 20 metallic spokes each $1 m$ long is rotated with a speed of $120 rpm$ in a plane perpendicular to a magnetic field of $0.4 G$. The induced emf between the axle and rim of the wheel will be. ($1 G = 10^{-4} T$) (2020-Covid)
Induced emf across the rotating rod/spoke is $e = \frac{1}{2} B \omega l^2$.
Here, $\omega = \frac{120 \times 2\pi}{60} = 4\pi rad s^{-1}$ and $B = 0.4 \times 10^{-4} T$.
$e = \frac{1}{2} \times (0.4 \times 10^{-4}) \times (4\pi) \times 1^2 = 0.8\pi \times 10^{-4} \approx 2.51 \times 10^{-4} V$.
24. A metallic rod of mass per unit length $0.5 kg m^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^\circ$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction $0.25 T$ is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is (2018)
For the rod to remain stationary on the incline, the forces along the plane must balance: $m g \sin\theta = I l B \cos\theta$.
Rearranging gives $I = \left(\frac{m}{l}\right) \frac{g \tan\theta}{B}$.
With $\frac{m}{l} = 0.5 kg m^{-1}$, $g = 9.8 m s^{-2}$, $\theta = 30^\circ$, and $B = 0.25 T$: $I = \frac{0.5 \times 9.8 \times \tan(30^\circ)}{0.25} = \frac{4.9}{0.25 \sqrt{3}} \approx 11.32 A$.