Magnetic Effect of Current

$\begin{aligned} &\text { Since velocity of light in terms of } \mu \& E \text { is }\\ &\begin{aligned} & V=\frac{1}{\sqrt{\mu \epsilon}}=\frac{1}{\sqrt{\mu_0 \mu_{\mathrm{r}}}} \times \frac{1}{\sqrt{\epsilon_0 \epsilon_{\mathrm{r}}}} \\ & =\frac{1}{\sqrt{\mu_{\mathrm{r}} \epsilon_{\mathrm{r}}}} \times \frac{1}{\sqrt{\mu_0 \epsilon_0}} \\ & =\frac{C}{\sqrt{\mu_{\mathrm{r}} \epsilon_{\mathrm{r}}}}=\frac{C}{\sqrt{\frac{10}{\pi} \times \frac{1}{0.0885}}} \\ & =\frac{C}{\sqrt{36}}=\frac{C}{6} \end{aligned} \end{aligned}$

$\begin{aligned} V & =\frac{C}{6} \\ C & =6 V \end{aligned}$

Velocity of light in vacuum is greater by 6 times the velocity of light in medium Answer is 6

Magnetic Field from Semi-Circular Segments:

The magnetic field at the center due to a semi-circular loop is given by:

$ B = \frac{\mu_0 I}{4R} $

Calculating $B_{R_1}$ and $B_{R_2}$:

For the semi-circle with radius $R_1$:

$ B_{R_1} = \frac{\mu_0 I}{4R_1} = \frac{4\pi \times 10^{-7} \times 12}{4 \times 6\pi} $

For the semi-circle with radius $R_2$:

$ B_{R_2} = \frac{\mu_0 I}{4R_2} = \frac{4\pi \times 10^{-7} \times 12}{4 \times 4\pi} $

Net Magnetic Field at the Center (O):

The total magnetic field $B_0$ at the center is the difference between these two fields (since they are in opposite directions):

$ B_0 = |B_{R_1} - B_{R_2}| $

Calculate:

$ B_0 = \frac{4\pi \times 10^{-7} \times 12}{4} \left(\frac{1}{4\pi} - \frac{1}{6\pi}\right) $

$ B_0 = 12\pi \times 10^{-7} \left(\frac{1}{12\pi}\right) $

$ B_0 = 1 \times 10^{-7} \text{ T} $

Value of $k$:

Given that the magnitude of the resultant magnetic field is $k \times 10^{-7} \text{ T}$, we find $k = 1$.

Thus, the value of $k$ is 1.

To calculate the magnetic force on a wire, we use the formula:

$ \mathrm{F} = \mathrm{I} \ell \mathrm{B} $

Where:

$ \mathrm{I} $ is the current in the wire (in amperes),

$ \ell $ is the length of the wire (in meters),

$\mathrm{B}$ is the magnetic field strength (in teslas).

Given that:

The current $\mathrm{I}$ is 8 A,

The length of the wire $\ell$ is 4.0 cm, which is 0.04 m (since 1 cm = 0.01 m),

The magnetic field strength $\mathrm{B}$ is 0.15 T,

Substituting these values into the formula gives:

$ \mathrm{F} = 8 \times 0.04 \times 0.15 $

Calculating this:

$ \mathrm{F} = 0.048 \, \mathrm{N} $

To convert this to millinewtons (mN), recall that $1 \, \mathrm{N} = 1000 \, \mathrm{mN}$. Therefore:

$ \mathrm{F} = 48 \, \mathrm{mN} $

Thus, the magnetic force on the wire is 48 mN.

We know, magnetic field due to solenoid, $B = {\mu _0}nI$

where, I = current, $n = {N \over L}$ = no. of turns per unit length

$ \Rightarrow B = {{{\mu _0}NI} \over L}$

$ \Rightarrow L = {{{\mu _0}NI} \over B} = {{4\pi \times {{10}^{ - 7}} \times 200 \times 0.29} \over {2.9 \times {{10}^{ - 4}}}}$

$ \Rightarrow L = 8\pi \times {10^{ - 2}}\,m$

$ \Rightarrow L = 8\pi \,cm$

Hence, answer = 8.

The problem involves an electron executing uniform circular motion inside a solenoid. The objective is to find the number of turns per meter in the solenoid.

Given:

Current in the solenoid, $ I = 1.5 \, \text{A} $

Time period of the electron's circular motion, $ T = 75 \, \text{ns} = 75 \times 10^{-9} \, \text{s} $

Mass of electron, $ m_e = 9 \times 10^{-31} \, \text{kg} $

Charge of electron, $ |q_e| = 1.6 \times 10^{-19} \, \text{C} $

Permeability of free space, $ \mu_0 = 4\pi \times 10^{-7} \, \text{N/A}^2 $

Explanation:

The time period $ T $ for a revolving charge in a magnetic field is given by:

$ T = \frac{2\pi m}{qB} $

The magnetic field $ B $ inside a solenoid is:

$ B = \mu_0 n I $

where $ n $ is the number of turns per meter. Thus, substituting $ B $ in the expression for $ T $, we get:

$ T = \frac{2\pi m}{q(\mu_0 n I)} $

Plugging in the known values:

$ 75 \times 10^{-9} = \frac{(2\pi)(9 \times 10^{-31})}{1.6 \times 10^{-19} \times 4\pi \times 10^{-7} \times n \times 1.5} $

Solving for $ n $, we find:

$ n = 250 $

Thus, the number of turns per meter in the solenoid is 250.

$ B = \frac{2 \sqrt{2}\,\mu_0 I}{\pi L} $

For a square loop of side length

$ L = \frac{1}{\sqrt{2}} \text{ m}, $

the perpendicular distance from the center to any side is

$ d = \frac{L}{2} = \frac{1}{2\sqrt{2}} \text{ m}. $

For a finite straight wire, the magnetic field at a point at distance $ d $ from the wire is given by the Biot–Savart expression

$ B_{\text{side}} = \frac{\mu_0 I}{4\pi d} \left(\sin\theta_1 + \sin\theta_2\right) $

where $ \theta_1 $ and $ \theta_2 $ are the angles between the extended wire and the line joining the ends of the wire with the observation point. For one side of the square, by symmetry, the midpoint of the side and the center of the square give

$ \tan\theta = \frac{L/2}{L/2} = 1 \quad\Rightarrow\quad \theta = 45^\circ. $

Thus,

$ \sin\theta_1 = \sin\theta_2 = \sin 45^\circ = \frac{1}{\sqrt{2}}. $

Then the field due to one side is

$ B_{\text{side}} = \frac{\mu_0 I}{4\pi \left(\frac{L}{2}\right)} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) = \frac{\mu_0 I}{4\pi \left(\frac{L}{2}\right)} \left(\frac{2}{\sqrt{2}}\right) = \frac{\mu_0 I \sqrt{2}}{2\pi L}. $

Since there are 4 sides with contributions that add vectorially in the same direction at the center, the total magnetic field is

$ B = 4 \cdot B_{\text{side}} = \frac{4 \mu_0 I \sqrt{2}}{2\pi L} = \frac{2 \sqrt{2}\,\mu_0 I}{\pi L}. $

Substitute

$ L = \frac{1}{\sqrt{2}}, $

to obtain

$ B = \frac{2 \sqrt{2}\,\mu_0 I}{\pi \left(\frac{1}{\sqrt{2}}\right)} = \frac{2 \sqrt{2}\,\mu_0 I \sqrt{2}}{\pi} = \frac{4\,\mu_0 I}{\pi}. $

Using the provided values

$ I = 5 \text{ A} \quad \text{and} \quad \mu_0 = 4\pi \times 10^{-7} \text{ T\,m/A}, $

the magnetic field becomes

$ B = \frac{4 \times \left(4\pi \times 10^{-7}\right) \times 5}{\pi} = \frac{80\pi \times 10^{-7}}{\pi} = 80 \times 10^{-7} \text{ T} = 8 \times 10^{-6} \text{ T}. $

Thus, the value of $ p $ is

$ p = 8. $

Let's break down the problem step by step:

When the proton moves undeflected in crossed electric and magnetic fields, the electric and magnetic forces balance each other. That is,

$qE = qvB,$

which simplifies to

$E = vB.$

After the electric field is switched off, the proton moves in a circular path under the action of the magnetic force. The magnetic force provides the required centripetal force:

$qvB = \frac{mv^2}{r}.$

Solving for the magnetic field $B$, we get:

$B = \frac{mv}{qr}.$

Now substitute this expression for $B$ back into the equilibrium condition:

$E = vB = v\left(\frac{mv}{qr}\right) = \frac{mv^2}{qr}.$

Plug in the given values:

Proton mass, $m = 1.6 \times 10^{-27} \, \text{kg}$

Speed, $v = 2 \times 10^5 \, \text{m/s}$

Radius, $r = 2 \, \text{cm} = 0.02 \, \text{m}$

Proton charge, $q = 1.6 \times 10^{-19} \, \text{C}$

Thus,

$E = \frac{(1.6 \times 10^{-27} \, \text{kg})(2 \times 10^5 \, \text{m/s})^2}{(1.6 \times 10^{-19} \, \text{C})(0.02 \, \text{m})}.$

Calculate the numerator:

$(2 \times 10^5)^2 = 4 \times 10^{10},$

So, $(1.6 \times 10^{-27}) \times (4 \times 10^{10}) = 6.4 \times 10^{-17}.$

Calculate the denominator:

$(1.6 \times 10^{-19}) \times (0.02) = 3.2 \times 10^{-21}.$

Now compute the electric field:

$E = \frac{6.4 \times 10^{-17}}{3.2 \times 10^{-21}} = 2 \times 10^4 \, \text{N/C}.$

The problem states that the magnitude of the electric field is $x \times 10^4 \, \text{N/C}.$ Since we found

$E = 2 \times 10^4 \, \text{N/C},$

it follows that

$x = 2.$

$\begin{aligned} & B=\frac{\mu_0(5)}{2 \pi \times .01}-\frac{\mu_0 4}{2 \pi \times 0.04} \\ & =-\frac{100 \mu_0}{4 \pi} \\ & =-100 \times 10^{-7} \\ & =-1 \times 10^{-5} \mathrm{~T} \end{aligned}$

Magnetic moment is defined as the product of the magnet's pole strength and the distance between the poles (also known as the magnetic length). When a magnetic strip is bent, its magnetic moment changes based on the new configuration.

Consider a straight magnetic strip with a magnetic moment of $44 \, \text{Am}^2$. If this strip is bent into a semicircular shape, we need to find the new effective magnetic moment.

The magnetic moment in a straight strip is given by:

$ M_{\text{straight}} = m \cdot l $

where:

• $m$ is the pole strength
• $l$ is the magnetic length

Given $M_{\text{straight}} = 44 \, \text{Am}^2$, let's now consider the strip bent into a semicircular shape.

When the strip is bent into a semicircle, the effective distance between the magnetic poles is the diameter of the semicircle. Let's denote the original length of the strip as $L$. In a straight line, this length $L$ is also the magnetic length. When bent into a semicircle, the length of the arc of the semicircle is still $L$.

The circumference of a full circle is given by:

$ C = 2\pi R $

Therefore, the length of the arc of a semicircle is:

$ L = \pi R $

Solving for $R$, we get:

$ R = \frac{L}{\pi} $

The diameter of the semicircle (which is the new effective magnetic length, $l_{\text{new}}$) is twice the radius:

$ l_{\text{new}} = 2R = 2 \cdot \frac{L}{\pi} = \frac{2L}{\pi} $

Now, the new magnetic moment $M_{\text{new}}$ is:

$ M_{\text{new}} = m \cdot l_{\text{new}} = m \cdot \frac{2L}{\pi} $

We know from the original magnetic strip:

$ M_{\text{straight}} = m \cdot L = 44 \, \text{Am}^2 $

Rewriting $m$ in terms of the known magnetic moment of the straight strip:

$ m = \frac{44}{L} $

Substituting $m$ into the new magnetic moment equation:

$ M_{\text{new}} = \frac{44}{L} \cdot \frac{2L}{\pi} $

Canceling out $L$ from the numerator and the denominator:

$ M_{\text{new}} = \frac{44 \cdot 2}{\pi} = \frac{88}{\pi} $

Given $\pi = \frac{22}{7}$, we substitute this value into the equation:

$ M_{\text{new}} = \frac{88 \cdot 7}{22} = 28 \, \text{Am}^2 $

Therefore, the magnetic moment of the strip when bent into a semicircular shape is $28 \, \text{Am}^2$.

Due to constant component of magnetic field $F = 0$

Due to variable component

$\begin{aligned} & F_1=0 \\ & \text { and, } F_2+F_3=0 \\ & \begin{aligned} \text { and, } F_4 & =\left(\mathrm{B}_0 4_x\right) i \mathrm{~L} \\ & =5 \times 4 \times 2 \times 2 \times 2 \\ & =160 \mathrm{~N} \end{aligned} \end{aligned}$

$\begin{aligned} & A = 36 \times 10^{-4} \mathrm{~m}^2 \\ & I= 6 \times 10^{-2} \mathrm{~m} \\ & =6 \mathrm{~cm} \\ V & =\frac{6 \mathrm{~cm}}{1 \mathrm{sec}}=6 \mathrm{~cm} / \mathrm{s} \\ \varepsilon & =B / \mathrm{vn}^2=0.5 \times \frac{6}{100} \times \frac{6}{100} \\ & =18 \times 10^{-4} \mathrm{~V} \\ E & =\frac{n^2 \varepsilon^2}{R} t=100 \times \frac{18 \times 18 \times 10^{-4} \times 10^{-4}}{10^2} \times 1 \\ & =324 \times 10^{-10} \times 10^2 \\ & =3.24 \times 10^{-6} \end{aligned}$

To solve this problem, we first need to understand that we want the electron to move along the same path in the presence of both electric and magnetic fields. This implies that the forces due to the electric field and magnetic field must balance each other.

The force on an electron due to the electric field is given by:

$F_E = eE$

where $e$ is the charge of the electron and $E$ is the electric field.

The force on an electron due to the magnetic field (Lorentz force) is given by:

$F_B = evB$

where $v$ is the velocity of the electron and $B$ is the magnetic field.

For the electron to move in a straight path, the forces due to the electric field and magnetic field must be equal in magnitude:

$eE = evB$

From this, we can solve for the electric field $E$:

$E = vB$

Next, we need to find the velocity $v$ of the electron. The kinetic energy (KE) of the electron is related to its velocity by the equation:

$KE = \frac{1}{2} mv^2$

Given the kinetic energy (KE) is $5 \, \text{eV}$, we first convert this energy into joules since the given constants are in SI units:

$5 \, \text{eV} = 5 \times 1.6 \times 10^{-19} \, \text{J} = 8 \times 10^{-19} \, \text{J}$

Now, solving for $v$:

$8 \times 10^{-19} = \frac{1}{2} \times 9 \times 10^{-31} \times v^2$

Rearrange to solve for $v^2$:

$v^2 = \frac{2 \times 8 \times 10^{-19}}{9 \times 10^{-31}}$

$v^2 = \frac{16 \times 10^{-19}}{9 \times 10^{-31}}$

$v^2 = \frac{16}{9} \times 10^{12}$

$v = \sqrt{\frac{16}{9} \times 10^{12}}$

$v = \frac{4}{3} \times 10^6 \, \text{m/s}$

Now we can find the electric field $E$. Using the value of the magnetic field $B$ given as $3 \, \mu \text{T} = 3 \times 10^{-6} \, \text{T}$:

$E = vB = \left(\frac{4}{3} \times 10^6 \, \text{m/s}\right) \times \left(3 \times 10^{-6} \, \text{T}\right)$

$E = \frac{4}{3} \times 3 \times 10^0 \, \text{N/C}$

$E = 4 \, \text{N/C}$

Therefore, the value of the electric field $E$ required for the electron to move along the same path is:

$\boxed{4 \, \text{N/C}}$

To find the work done in turning the coil through $90^{\circ}$, we first need to understand the concept of torque on a current-carrying loop in a magnetic field and how work done relates to the change in potential energy of the system. The potential energy (U) of a magnetic dipole in a magnetic field is given by:

$U = - \vec{M} \cdot \vec{B}$

where:

• $\vec{M}$ is the magnetic moment of the coil, and
• $\vec{B}$ is the magnetic field.

For a coil with $N$ turns, carrying current $I$, and with an area $A$, the magnetic moment $\vec{M}$ is defined as:

$M = NI \cdot A$

Given that the coil has $100$ turns, carries a current of $1 \, \mathrm{mA} = 1 \times 10^{-3} \, \mathrm{A}$, and the area of the coil is $5 \times 10^{-3} \, \mathrm{m}^2$, we can calculate its magnetic moment as follows:

$M = 100 \cdot 1 \times 10^{-3} \cdot 5 \times 10^{-3} = 0.5 \times 10^{-3} \, \mathrm{Am}^2$

Since the coil is initially placed such that its plane is perpendicular to the magnetic field (i.e., the angle $ \theta = 0^{\circ} $), and then it is turned through $90^{\circ}$, the initial and final angles ($\theta_i$ and $\theta_f$) of the coil with respect to the magnetic field are $0^{\circ}$ and $90^{\circ}$ respectively. This means the initial potential energy ($U_i$) and final potential energy ($U_f$) of the system are:

$U_i = - M B \cos(\theta_i)$

$U_f = - M B \cos(\theta_f)$

Given that $B = 0.20 \, \mathrm{T}$, $\theta_i = 0^{\circ} \, (\cos(0) = 1)$, and $\theta_f = 90^{\circ} \, (\cos(90^{\circ}) = 0)$, the potential energies are:

$U_i = - 0.5 \times 10^{-3} \cdot 0.20 \cdot 1 = - 1 \times 10^{-4} \, \mathrm{J}$

$U_f = - 0.5 \times 10^{-3} \cdot 0.20 \cdot 0 = 0 \, \mathrm{J}$

The work done ($W$) is equal to the change in potential energy:

$W = U_f - U_i$

$W = 0 - (- 1 \times 10^{-4}) = 1 \times 10^{-4} \, \mathrm{J}$

Therefore, the work done in turning the coil through $90^{\circ}$ is $100 \mu \mathrm{J}$.

$\begin{aligned} & W=U_f-U_i=(-M B \cos 90)-(-M B \cos 0) \\ & \Rightarrow W=M B=N i A B=5 \mu \mathrm{J} \end{aligned}$

The magnetic field inside a solenoid can be calculated using the formula:

$B = \mu_0 n I$

where:

• $B$ is the magnetic field in teslas (T),
• $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \mathrm{~Tm/A}$),
• $n$ is the number of turns per unit length of the solenoid (turns/m),
• $I$ is the current in amperes (A).

Given:

• The magnetic field $B = 6.28 \times 10^{-3} \mathrm{~T}$,
• The current $I = 5 \mathrm{~A}$,
• The length of the solenoid $L = 0.5 \mathrm{~m}$,

First, let's calculate the number of turns per unit length $n$, which is $n = \frac{m}{L}$ where $m$ is the total number of turns and $L$ is the length of the solenoid.

Rearrange the formula for $B$ to solve for $m$:

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

Therefore,

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

Substituting the values we have:

$m = \frac{(6.28 \times 10^{-3} \mathrm{T}) (0.5 \mathrm{m})}{(4\pi \times 10^{-7} \mathrm{Tm/A}) (5 \mathrm{A})}$

$m = \frac{6.28 \times 10^{-3} \times 0.5}{4\pi \times 10^{-7} \times 5}$

$m = \frac{6.28 \times 0.5 \times 10^{-3}}{20\pi \times 10^{-7}}$

$m = \frac{3.14 \times 10^{-3}}{20 \pi \times 10^{-7}}$

$m = \frac{3.14 \times 10^{-3}}{20 \times 3.14 \times 10^{-7}}$

$m = \frac{1}{20 \times 10^{-4}}$

$m = \frac{1 \times 10^4}{20}$

$m = 500$

Therefore, the value of $m$ is 500 turns.

$\begin{aligned} & i L B=m g \text { and } L=\frac{A R}{\rho} \\ & \begin{aligned} \therefore B & =\frac{m g \rho}{i A R} \\ & =\frac{0.5 \times 10 \times 2 \times 10^{-6}}{2 \times 10 \times 10^{-6} \times 1} \\ & =0.5 \mathrm{~T} \end{aligned} \end{aligned}$

At point $A$

$B_A=\frac{\mu_0 I}{2 \pi r}+\frac{\mu_0(2 I)}{2 \pi(3 r)}$

At point $C$

$\begin{aligned} & B_C=\frac{\mu_0 I}{2 \pi(3 r)}+\frac{\mu_0(2 I)}{r} \\ & \Rightarrow \frac{B_A}{B_C}=\frac{5}{7} \end{aligned}$

Both end having same potential, so potential difference between end will be zero.