Electromagnetic Induction

The induced charge in the coil is given by Faraday's law:$$Q=\frac{N\cdot A\cdot B}{R}\cdot \mathrm{\Delta }t$$
$N=100$ is the number of turns,
$A=0.05{\text{m}}^2$ is the area of each turn,
$B=90\text{mT}=0.09\text{T}$is the magnetic field strength,
$R=1.5\mathrm{\Omega }$is the total resistance of the coil.

Substitute these values into the formula:$$Q=\frac{100\cdot 0.05\cdot 0.09}{1.5}=0.30\text{C}$$

The self inductance L of a solenoid of length I and area of cross-section A, with a fixed number of turns N increases as \(\underline{\mathbf{l}}\) decreases and A increases.

Explanation:
The self inductance L of a solenoid depends on various factor like geometry and magnetic permeability of the core material.

\(\mathrm{L}=\mu_{\mathrm{r}} \mu_0 \mathrm{~N}^2 \mathrm{Al}\)

Where \(\mathrm{n}=\mathrm{N} / \mathrm{l}\) (no. of turns per unit length)
1. No. of turns: Larger the number of turns in solenoid, larger is its self inductance.
2. Area of cross-section: Larger the area of cross-section of the solenoid, larger is its self inductance.
3. Permeability of the core material. The self inductance of a solenoid increases \(\mu \mathrm{r}\) times if it is wound over an iron core of relative permeability \(\mu_r\).

The long solenoid of cross-sectional area A and length 1 , having A turns, filled inside of the solenoid with a material of relative permeability (e.g., soft iron, which has a high value of relative permeability) then its self inductance is \(L=\mu_r \mu_0 N^2 A / l\)

So, the self inductance \(L\)of a solenoid increases as 1 decreases and \(A\) increases because \(L\) is directly proportional to the area and inversely proportional to length.

Important point: The self and mutual inductance of capacitance and resistance depend on the geometry of the devices as well as permittivity/ permeability of the medium.

Use Faraday's law: $\mathcal{E}=N A\frac{\Delta B}{\Delta t}$. Area $A$ and $N$ cancel in the ratio.

$$\frac{e_1}{e_2}=\frac{\Delta B/\Delta t_1}{\Delta B/\Delta t_2} =\frac{(30\times10^{-3})/(20\times10^{-3})}{(30\times10^{-3})/(40\times10^{-3})} =\frac{30/20}{30/40}=\frac{40}{20}=2.$$

So $e_1/e_2=2.$

Calculate the induced emf using the formula: \(\operatorname{emf}=-L \frac{d i}{d t}=-15 \times 10^{-3} \frac{4}{0.004}=-15 \times 10^{-3} \times 1000=\) -15 V . The negative sign indicates the direction of the induced emf, but we are interested in the magnitude, which is 15 V .

we use the formula for the number of turns per unit length of a solenoid:

where:

(Inductance)

(Length of solenoid)

(Radius)

(Cross-sectional area)

(Turns per unit length)

Rearranging for N :

Substituting values:

$\varphi =BAN·\cos \left(\omega t\right)$

$\epsilon =\frac{-d\varphi }{dt}=BA\omega N·\sin \left(\omega t\right)$

When B is parallel to plane, $\omega t=\frac{\pi }{2}$

$\Rightarrow \varphi =0,\epsilon =BA\omega N$

The given problem involves a rotating disc in the presence of a uniform magnetic field, requiring us to determine the potential difference between the center and the rim.

• Angular velocity: $\omega = 10\pi$

• Magnetic field: $B = 0.4$ (pointing perpendicular to the plane)

• Radius of the disc: $R = 20$

• Formula to be used: The induced potential difference (emf) between the center and the rim of a rotating disc in a magnetic field is given by:

  $$V = \frac{1}{2} B \omega R^2$$

$$V = \frac{1}{2} \times (0.4) \times (10\pi) \times (0.2)^2$$

$$V = \frac{1}{2} \times 0.4 \times 10\pi \times 0.04$$

$$V = \frac{1}{2} \times 0.16\pi$$

$$V = 0.08\pi$$

Using $\pi \approx 3.14$

$$V = 0.08 \times 3.14$$

$$V \approx 0.251 \text{ V}$$

The potential difference between the center and the rim of the disc is approximately 0.251 V.

The induced charge in the coil is given by Faraday's law:$$Q=\frac{N\cdot A\cdot B}{R}\cdot \mathrm{\Delta }t$$
$N=100$ is the number of turns,
$A=0.05{\text{m}}^2$ is the area of each turn,
$B=90\text{mT}=0.09\text{T}$is the magnetic field strength,
$R=1.5\mathrm{\Omega }$is the total resistance of the coil.

Substitute these values into the formula:$$Q=\frac{100\cdot 0.05\cdot 0.09}{1.5}=0.30\text{C}$$

we use the formula for the number of turns per unit length of a solenoid:

where:

(Inductance)

(Length of solenoid)

(Radius)

(Cross-sectional area)

(Turns per unit length)

Rearranging for N :

Substituting values:

Due to loop A

$$\begin{gathered} B_A = \frac{{\mu }_{o}i}{2R} \\ {\phi }_B = B_A{A}_B = \frac{{\mu }_{o}i}{2R}\pi \frac{R^2}{{\left(20\right)}^2} \\ {\phi }_B = \frac{{\mu }_o\pi iR}{800} \end{gathered}$$

\[
\text{Correct option is: (B) }~2866
\]

\[
\text{We use the formula for the number of turns per unit length of a solenoid:}
\]

\[
L = \mu_0 n^2 A l
\]

\[
\begin{aligned}
\text{where:} \quad
&L = 0.016~\mathrm{H} &&\text{(Inductance)} \\
&l = 0.81~\mathrm{m} &&\text{(Length of solenoid)} \\
&r = 0.02~\mathrm{m} &&\text{(Radius)} \\
&A = \pi r^2 &&\text{(Cross-sectional area)} \\
&n = \frac{N}{l} &&\text{(Turns per unit length)}
\end{aligned}
\]

\[
\text{Rearranging for } N:
\quad
N = \sqrt{\frac{L}{\mu_0 A}} \times l
\]

\[
\begin{aligned}
N &= \sqrt{\frac{0.016}{\left(4\pi \times 10^{-7}\right)\left(\pi \times (0.02)^2\right)} \times 0.81} \\[6pt]
&= \sqrt{\frac{16 \times 81}{100 \times 1000 \times 10^{-11} \times 16\pi^2}} \\[6pt]
&= \frac{4 \times 9 \times 1000}{4\pi} \\[6pt]
&= \frac{9000}{\pi} \\[6pt]
&\Rightarrow N \approx 2866
\end{aligned}
\]

The magnetic flux ΦB linked with the coil is given by:${\varphi }_B=BACos\theta$

The area of the coil is:A=π$r^2$=π($0.1^2$)=0.0314

The magnetic field B is the dot product of the vector B and the unit vector normal to the surface of the coil:→B⋅$\hat{n}$= (1.0i+0.5j)⋅(0.6i+0.8j)=1.0×0.6+0.5×0.8=0.6+0.4=1.0mT.

Thus, the magnetic flux is:${\varphi }_B$=1.0×0.0314=0.0314mWb=31.4μWb.

(a) Using the expression of magnetic flux.

The self inductance L of a solenoid of length I and area of cross-section A, with a fixed number of turns N increases as \(\underline{\mathbf{l}}\) decreases and A increases.

Explanation:
The self inductance L of a solenoid depends on various factor like geometry and magnetic permeability of the core material.

\(\mathrm{L}=\mu_{\mathrm{r}} \mu_0 \mathrm{~N}^2 \mathrm{Al}\)

Where \(\mathrm{n}=\mathrm{N} / \mathrm{l}\) (no. of turns per unit length)
1. No. of turns: Larger the number of turns in solenoid, larger is its self inductance.
2. Area of cross-section: Larger the area of cross-section of the solenoid, larger is its self inductance.
3. Permeability of the core material. The self inductance of a solenoid increases \(\mu \mathrm{r}\) times if it is wound over an iron core of relative permeability \(\mu_r\).

The long solenoid of cross-sectional area A and length 1 , having A turns, filled inside of the solenoid with a material of relative permeability (e.g., soft iron, which has a high value of relative permeability) then its self inductance is \(L=\mu_r \mu_0 N^2 A / l\)

So, the self inductance \(L\)of a solenoid increases as 1 decreases and \(A\) increases because \(L\) is directly proportional to the area and inversely proportional to length.

Important point: The self and mutual inductance of capacitance and resistance depend on the geometry of the devices as well as permittivity/ permeability of the medium.

Use Faraday's law: $\mathcal{E}=N A\frac{\Delta B}{\Delta t}$. Area $A$ and $N$ cancel in the ratio.

$$\frac{e_1}{e_2}=\frac{\Delta B/\Delta t_1}{\Delta B/\Delta t_2} =\frac{(30\times10^{-3})/(20\times10^{-3})}{(30\times10^{-3})/(40\times10^{-3})} =\frac{30/20}{30/40}=\frac{40}{20}=2.$$

So $e_1/e_2=2.$

We are given:

• Radius of solenoid = $r = 10$cm (not needed in calculations)
• Current = $I = 0.29$ A
• Total number of turns = $N = 200$
• Magnetic field inside the solenoid = $B = 2.9 \times 10^{-4}$

We need to determine the length of the solenoid ($L$).

The magnetic field inside a solenoid is given by:

$$B = \mu_0 n I$$

where:

• $\mu_0$​ is the permeability of free space, $\mu_0 = 4\pi \times 10^{-7}$

• $n$n is the number of turns per unit length, given by:

  $$n = \frac{N}{L}$$​

• $I$ is the current.

Rewriting the equation:

$$B = \mu_0 \frac{N}{L} I$$

Solving for $L$:

$$L = \frac{\mu_0 N I}{B}$$​

$$L = \frac{(4\pi \times 10^{-7}) (200) (0.29)}{2.9 \times 10^{-4}}$$

​ $$L = \frac{(4\pi \times 10^{-7} \times 200 \times 0.29)}{2.9 \times 10^{-4}}$$

$L=8\pi \text{ cm}$

$$\boxed{B} \quad 8\pi \text{ cm}$$

The induced charge in the coil is given by Faraday's law:$$Q=\frac{N\cdot A\cdot B}{R}\cdot \mathrm{\Delta }t$$
$N=100$ is the number of turns,
$A=0.05{\text{m}}^2$ is the area of each turn,
$B=90\text{mT}=0.09\text{T}$is the magnetic field strength,
$R=1.5\mathrm{\Omega }$is the total resistance of the coil.

Substitute these values into the formula:$$Q=\frac{100\cdot 0.05\cdot 0.09}{1.5}=0.30\text{C}$$

Calculate the induced emf using the formula: \(\operatorname{emf}=-L \frac{d i}{d t}=-15 \times 10^{-3} \frac{4}{0.004}=-15 \times 10^{-3} \times 1000=\) -15 V . The negative sign indicates the direction of the induced emf, but we are interested in the magnitude, which is 15 V .

\[
\text{Correct option is: (B) }~2866
\]

\[
\text{We use the formula for the number of turns per unit length of a solenoid:}
\]

\[
L = \mu_0 n^2 A l
\]

\[
\begin{aligned}
\text{where:} \quad
&L = 0.016~\mathrm{H} &&\text{(Inductance)} \\
&l = 0.81~\mathrm{m} &&\text{(Length of solenoid)} \\
&r = 0.02~\mathrm{m} &&\text{(Radius)} \\
&A = \pi r^2 &&\text{(Cross-sectional area)} \\
&n = \frac{N}{l} &&\text{(Turns per unit length)}
\end{aligned}
\]

\[
\text{Rearranging for } N:
\quad
N = \sqrt{\frac{L}{\mu_0 A}} \times l
\]

\[
\begin{aligned}
N &= \sqrt{\frac{0.016}{\left(4\pi \times 10^{-7}\right)\left(\pi \times (0.02)^2\right)} \times 0.81} \\[6pt]
&= \sqrt{\frac{16 \times 81}{100 \times 1000 \times 10^{-11} \times 16\pi^2}} \\[6pt]
&= \frac{4 \times 9 \times 1000}{4\pi} \\[6pt]
&= \frac{9000}{\pi} \\[6pt]
&\Rightarrow N \approx 2866
\end{aligned}
\]

\(\varepsilon = -\frac{\Delta \phi}{\Delta t}\)

\(= -\frac{0 - (4 \times (2.5 \times 2) \cos 60^{\circ})}{10} = +1V\)

we use the formula for the number of turns per unit length of a solenoid:

where:

(Inductance)

(Length of solenoid)

(Radius)

(Cross-sectional area)

(Turns per unit length)

Rearranging for N :

Substituting values:

We are given:

• Radius of solenoid = $r = 10$cm (not needed in calculations)
• Current = $I = 0.29$ A
• Total number of turns = $N = 200$
• Magnetic field inside the solenoid = $B = 2.9 \times 10^{-4}$

We need to determine the length of the solenoid ($L$).

The magnetic field inside a solenoid is given by:

$$B = \mu_0 n I$$

where:

• $\mu_0$​ is the permeability of free space, $\mu_0 = 4\pi \times 10^{-7}$

• $n$n is the number of turns per unit length, given by:

  $$n = \frac{N}{L}$$​

• $I$ is the current.

Rewriting the equation:

$$B = \mu_0 \frac{N}{L} I$$

Solving for $L$:

$$L = \frac{\mu_0 N I}{B}$$​

$$L = \frac{(4\pi \times 10^{-7}) (200) (0.29)}{2.9 \times 10^{-4}}$$

​ $$L = \frac{(4\pi \times 10^{-7} \times 200 \times 0.29)}{2.9 \times 10^{-4}}$$

$L=8\pi \text{ cm}$

$$\boxed{B} \quad 8\pi \text{ cm}$$

$$\begin{gathered} \epsilon = -NA\frac{dB}{dt}= NA{B}_0\frac{2\pi }{T}\sin \left(\frac{2\pi }{T}t\right) \\ \sin \left(\frac{2\pi }{T}t\right) = 1 \\ condition \sin \left(\frac{2\pi }{T}t\right) = 1 is met, when \\ \frac{2\pi }{T}t = \frac{(2n-1)\pi }{2} for any integer n (1,2,3,...) \\ t = \frac{(2n-1)T}{4} \end{gathered}$$

the magnitude of the induced emf is maximum at times such $t =\frac{T}{4}, \frac{3T}{4},\frac{5T}{4},...$

Use Faraday's law: $\mathcal{E}=N A\frac{\Delta B}{\Delta t}$. Area $A$ and $N$ cancel in the ratio.

$$\frac{e_1}{e_2}=\frac{\Delta B/\Delta t_1}{\Delta B/\Delta t_2} =\frac{(30\times10^{-3})/(20\times10^{-3})}{(30\times10^{-3})/(40\times10^{-3})} =\frac{30/20}{30/40}=\frac{40}{20}=2.$$

So $e_1/e_2=2.$

Use Faraday's law: $\mathcal{E}=N A\frac{\Delta B}{\Delta t}$. Area $A$ and $N$ cancel in the ratio.

$$\frac{e_1}{e_2}=\frac{\Delta B/\Delta t_1}{\Delta B/\Delta t_2} =\frac{(30\times10^{-3})/(20\times10^{-3})}{(30\times10^{-3})/(40\times10^{-3})} =\frac{30/20}{30/40}=\frac{40}{20}=2.$$

So $e_1/e_2=2.$

\[
\text{Correct option is: (B) }~2866
\]

\[
\text{We use the formula for the number of turns per unit length of a solenoid:}
\]

\[
L = \mu_0 n^2 A l
\]

\[
\begin{aligned}
\text{where:} \quad
&L = 0.016~\mathrm{H} &&\text{(Inductance)} \\
&l = 0.81~\mathrm{m} &&\text{(Length of solenoid)} \\
&r = 0.02~\mathrm{m} &&\text{(Radius)} \\
&A = \pi r^2 &&\text{(Cross-sectional area)} \\
&n = \frac{N}{l} &&\text{(Turns per unit length)}
\end{aligned}
\]

\[
\text{Rearranging for } N:
\quad
N = \sqrt{\frac{L}{\mu_0 A}} \times l
\]

\[
\begin{aligned}
N &= \sqrt{\frac{0.016}{\left(4\pi \times 10^{-7}\right)\left(\pi \times (0.02)^2\right)} \times 0.81} \\[6pt]
&= \sqrt{\frac{16 \times 81}{100 \times 1000 \times 10^{-11} \times 16\pi^2}} \\[6pt]
&= \frac{4 \times 9 \times 1000}{4\pi} \\[6pt]
&= \frac{9000}{\pi} \\[6pt]
&\Rightarrow N \approx 2866
\end{aligned}
\]

The self inductance L of a solenoid of length I and area of cross-section A, with a fixed number of turns N increases as \(\underline{\mathbf{l}}\) decreases and A increases.

Explanation:
The self inductance L of a solenoid depends on various factor like geometry and magnetic permeability of the core material.

\(\mathrm{L}=\mu_{\mathrm{r}} \mu_0 \mathrm{~N}^2 \mathrm{Al}\)

Where \(\mathrm{n}=\mathrm{N} / \mathrm{l}\) (no. of turns per unit length)
1. No. of turns: Larger the number of turns in solenoid, larger is its self inductance.
2. Area of cross-section: Larger the area of cross-section of the solenoid, larger is its self inductance.
3. Permeability of the core material. The self inductance of a solenoid increases \(\mu \mathrm{r}\) times if it is wound over an iron core of relative permeability \(\mu_r\).

The long solenoid of cross-sectional area A and length 1 , having A turns, filled inside of the solenoid with a material of relative permeability (e.g., soft iron, which has a high value of relative permeability) then its self inductance is \(L=\mu_r \mu_0 N^2 A / l\)

So, the self inductance \(L\)of a solenoid increases as 1 decreases and \(A\) increases because \(L\) is directly proportional to the area and inversely proportional to length.

Important point: The self and mutual inductance of capacitance and resistance depend on the geometry of the devices as well as permittivity/ permeability of the medium.

(a) Using the expression of magnetic flux.

The magnetic flux ΦB linked with the coil is given by:${\varphi }_B=BACos\theta$

The area of the coil is:A=π$r^2$=π($0.1^2$)=0.0314

The magnetic field B is the dot product of the vector B and the unit vector normal to the surface of the coil:→B⋅$\hat{n}$= (1.0i+0.5j)⋅(0.6i+0.8j)=1.0×0.6+0.5×0.8=0.6+0.4=1.0mT.

Thus, the magnetic flux is:${\varphi }_B$=1.0×0.0314=0.0314mWb=31.4μWb.

Self inductance of coil
$L=\frac{{\mu }_0{N}^{2}A}{2\pi R}$

B is incorrect because Self-inductance depends on the permeability of the medium through which the coil is wound. Higher permeability increases inductance

$\varphi =BAN·\cos \left(\omega t\right)$

$\epsilon =\frac{-d\varphi }{dt}=BA\omega N·\sin \left(\omega t\right)$

When B is parallel to plane, $\omega t=\frac{\pi }{2}$

$\Rightarrow \varphi =0,\epsilon =BA\omega N$

Self Inductance depends on the geometry of the conductor and the property of material medium.

Calculate the induced emf using the formula: \(\operatorname{emf}=-L \frac{d i}{d t}=-15 \times 10^{-3} \frac{4}{0.004}=-15 \times 10^{-3} \times 1000=\) -15 V . The negative sign indicates the direction of the induced emf, but we are interested in the magnitude, which is 15 V .

The magnetic flux ΦB linked with the coil is given by:${\varphi }_B=BACos\theta$

The area of the coil is:A=π$r^2$=π($0.1^2$)=0.0314

The magnetic field B is the dot product of the vector B and the unit vector normal to the surface of the coil:→B⋅$\hat{n}$= (1.0i+0.5j)⋅(0.6i+0.8j)=1.0×0.6+0.5×0.8=0.6+0.4=1.0mT.

Thus, the magnetic flux is:${\varphi }_B$=1.0×0.0314=0.0314mWb=31.4μWb.