Static EMI (Time Varying Magnetic Field) Questions in English

(D) The magnetic field is given by $\vec{B} = B_0 r t \hat{k}$.
According to Faraday's law of induction in integral form,$\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}$.
For a circular path of radius $r$ centered on the $z$-axis,the line integral of the electric field is $E(2\pi r)$.
The magnetic flux $\phi_B$ through the circular area of radius $r$ is $\int_0^r B(r') (2\pi r') dr'$.
Substituting $B(r') = B_0 r' t$,we get $\phi_B = \int_0^r (B_0 r' t) (2\pi r') dr' = 2\pi B_0 t \int_0^r r'^2 dr' = 2\pi B_0 t \frac{r^3}{3}$.
Now,differentiating with respect to time $t$: $\frac{d\phi_B}{dt} = \frac{d}{dt} \left( \frac{2\pi B_0 t r^3}{3} \right) = \frac{2\pi B_0 r^3}{3}$.
Equating the magnitudes: $E(2\pi r) = \frac{2\pi B_0 r^3}{3}$.
Solving for $E$,we get $E = \frac{B_0 r^2}{3}$.

(A) According to Faraday's law of electromagnetic induction,the line integral of the electric field is equal to the negative rate of change of magnetic flux: $\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi}{dt}$.
For a long solenoid,the magnetic field inside is $B = \mu_0 n I$. The magnetic flux through a circular loop of radius $r$ is $\phi = B \cdot A$.
Case $1$: For $r < a$,the flux is $\phi = (\mu_0 n I)(\pi r^2)$.
Applying Faraday's law: $E(2 \pi r) = -\frac{d}{dt}(\mu_0 n I \pi r^2) = -\mu_0 n \pi r^2 \frac{dI}{dt}$.
Thus,$E = -\frac{\mu_0 n r}{2} \frac{dI}{dt}$. The magnitude is $|E| = \frac{\mu_0 n r \dot{I}}{2}$.
Case $2$: For $r > a$,the flux is only through the cross-section of the solenoid,so $\phi = (\mu_0 n I)(\pi a^2)$.
Applying Faraday's law: $E(2 \pi r) = -\frac{d}{dt}(\mu_0 n I \pi a^2) = -\mu_0 n \pi a^2 \frac{dI}{dt}$.
Thus,$E = -\frac{\mu_0 n a^2}{2 r} \frac{dI}{dt}$. The magnitude is $|E| = \frac{\mu_0 n a^2 \dot{I}}{2 r}$.

(A, B, C, D) For $r < R$,using Faraday's law of induction: $\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}$.
$E(2\pi r) = \frac{d}{dt}[(4t^2 - 2t + 6) \pi r^2] = (8t - 2) \pi r^2$.
Thus,$E = \frac{(8t - 2)r}{2} = (4t - 1)r$. Since $E \propto r$,the electric field varies linearly with $r$ for $r < R$.
For $r > R$,the magnetic flux is confined to the region $r \le R$: $\oint \vec{E} \cdot d\vec{l} = -\frac{d}{dt}[B \cdot \pi R^2]$.
$E(2\pi r) = (8t - 2) \pi R^2 \implies E = \frac{(8t - 2)R^2}{2r}$.
At $r = 2R$ and $t = 2 \text{ s}$:
$E = \frac{(8(2) - 2)R^2}{2(2R)} = \frac{14R^2}{4R} = \frac{7R}{2}$.
The acceleration $a = \frac{Eq}{m} = \frac{7qR}{2m}$.
Induced electric fields are non-conservative and form closed loops. By Lenz's law,since $\vec{B}$ is increasing for $t > 0.25 \text{ s}$,the induced field opposes the change,resulting in a clockwise direction for a positive charge.

(D) The magnetic flux $\Phi$ through the loop is given by $\Phi = B A \cos \theta$.
Here,$A = (2 \text{ cm})^2 = (0.02 \text{ m})^2 = 4 \times 10^{-4} \text{ m}^2$ and $\theta = 60^{\circ}$.
Substituting the values: $\Phi = (0.4 \sin(300t)) \times (4 \times 10^{-4}) \times \cos(60^{\circ})$.
Since $\cos(60^{\circ}) = 0.5$,we get $\Phi = 0.4 \times 4 \times 10^{-4} \times 0.5 \times \sin(300t) = 0.8 \times 10^{-4} \sin(300t) \text{ Wb}$.
The induced emf $\epsilon$ is given by Faraday's law: $\epsilon = -\frac{d\Phi}{dt}$.
$\epsilon = -\frac{d}{dt} [0.8 \times 10^{-4} \sin(300t)] = -0.8 \times 10^{-4} \times 300 \times \cos(300t) = -0.024 \cos(300t) \text{ V}$.
The maximum induced emf is the amplitude of this expression,which is $\epsilon_{max} = 0.024 \text{ V}$.
Converting to millivolts: $0.024 \text{ V} = 24 \text{ mV}$.

(A) The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = (2t^2 + 2t + 3) \cdot \pi r^2$.
Given $r = 20 \text{ cm} = 0.2 \text{ m}$,so $A = \pi (0.2)^2 = 0.04\pi \text{ m}^2$.
Thus,$\phi = 0.04\pi (2t^2 + 2t + 3) \text{ Wb}$.
The induced emf $\varepsilon$ is given by Faraday's law: $\varepsilon = |-\frac{d\phi}{dt}|$.
$\frac{d\phi}{dt} = 0.04\pi (4t + 2)$.
At $t = 3 \text{ s}$,$\frac{d\phi}{dt} = 0.04\pi (4(3) + 2) = 0.04\pi (14) = 0.56\pi \text{ V}$.
The induced current $I$ is $I = \frac{\varepsilon}{R} = \frac{0.56\pi}{2} = 0.28\pi \text{ A}$.
Using $\pi \approx \frac{22}{7}$,$I = 0.28 \times \frac{22}{7} = 0.04 \times 22 = 0.88 \text{ A}$.
Given $I = \frac{\alpha}{50}$,we have $0.88 = \frac{\alpha}{50}$.
Therefore,$\alpha = 0.88 \times 50 = 44$.