Ampere’s circuital law and its application (Solenoid and Toroid) Questions in English

(D) Given:
Current in the long wire, $I = 18 \,A$.
Magnetic field due to the solenoid, $B_1 = 8.0 \times 10^{-3} \,T$ (directed along the axis).
Distance of point $P$ from the axis, $r = 0.6 \,mm = 0.6 \times 10^{-3} \,m$.
The magnetic field due to the long current-carrying wire at a distance $r$ is given by:
$B_2 = \frac{\mu_0 I}{2 \pi r} = \frac{2 \times 10^{-7} \times 18}{0.6 \times 10^{-3}} = \frac{36 \times 10^{-7}}{0.6 \times 10^{-3}} = 60 \times 10^{-4} \,T = 6 \times 10^{-3} \,T$.
The magnetic field $B_1$ due to the solenoid is along the axis, and the magnetic field $B_2$ due to the wire is tangential to the circle of radius $r$ around the wire. Thus, $B_1$ and $B_2$ are perpendicular to each other.
The resultant magnetic field $B$ is:
$B = \sqrt{B_1^2 + B_2^2} = \sqrt{(8 \times 10^{-3})^2 + (6 \times 10^{-3})^2} \,T$
$B = \sqrt{64 \times 10^{-6} + 36 \times 10^{-6}} \,T = \sqrt{100 \times 10^{-6}} \,T = 10 \times 10^{-3} \,T$.

(C) The mean radius $r_m$ of the toroid is calculated as:
$r_m = \frac{0.24 + 0.26}{2} = 0.25 \ m$
The magnetic field $B$ inside a toroid is given by the formula:
$B = \mu_0 n I = \frac{\mu_0 N I}{2 \pi r_m}$
Substituting the given values:
$N = 2500$, $I = 10 \ A$, $r_m = 0.25 \ m$, and $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$
$B = \frac{4 \pi \times 10^{-7} \times 2500 \times 10}{2 \pi \times 0.25}$
$B = \frac{2 \times 10^{-7} \times 25000}{0.25} = \frac{5 \times 10^{-3}}{0.25} = 20 \times 10^{-3} = 2 \times 10^{-2} \ T$

(C) For a thin hollow cylinder of radius $R$ carrying a uniform current $i$ along its length:
$1$. Inside the cylinder $(d < R)$, the magnetic field $B$ is zero because the enclosed current is zero.
$2$. Outside the cylinder $(d \geq R)$, the magnetic field $B$ is given by Ampere's Law as $B = \frac{\mu_{0} i}{2 \pi d}$. This shows that $B \propto \frac{1}{d}$.
Therefore, the graph of $B$ versus $d$ will show $B = 0$ for $d < R$ and a hyperbolic decay for $d \geq R$. This corresponds to the plot shown in figure $C$.

(A) For a long cylindrical conductor of radius $R$ carrying a current $I$ distributed uniformly, the magnetic field $B$ at a distance $r$ from the axis is given by Ampere's circuital law:
$1$. Inside the conductor $(r < R)$: $B = \frac{\mu_0 I r}{2 \pi R^2}$. At the axis $(r = 0)$, $B = 0$, which is the minimum value.
$2$. At the surface $(r = R)$: $B = \frac{\mu_0 I}{2 \pi R}$, which is the maximum value.
$3$. Outside the conductor $(r > R)$: $B = \frac{\mu_0 I}{2 \pi r}$, which decreases as $r$ increases.
Thus, the magnetic field is minimum at the axis of the conductor (statement $D$). Statement $A, B, C, E$ are incorrect. Therefore, only statement $D$ is correct.

(A) The magnetic field $\vec{B}$ inside a long solenoid is directed along its axis.
Since the particle is released from rest and moves along the axis of the solenoid, its velocity vector $\vec{v}$ is always parallel or anti-parallel to the magnetic field vector $\vec{B}$.
The magnetic force on a moving charge is given by $\vec{F}_{B} = Q(\vec{v} \times \vec{B})$.
Since $\vec{v}$ is parallel to $\vec{B}$, the cross product $\vec{v} \times \vec{B} = 0$, hence $\vec{F}_{B} = 0$.
The only force acting on the particle is the gravitational force $\vec{F}_{g} = m\vec{g}$ acting downwards.
Therefore, the net force $\vec{F}_{net} = m\vec{g}$.
According to Newton's second law, $m\vec{a} = m\vec{g}$, which gives $\vec{a} = \vec{g}$.
Thus, the acceleration of the particle is $a = g$.

(B) The magnetic field $B$ in a solenoid with a core is given by the formula $B = \mu_r \mu_0 H$, where $H$ is the magnetic intensity.
Given values are $B = 1.0 \text{ T}$, $\mu_r = 400$, and $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$.
Rearranging the formula to solve for $H$, we get $H = \frac{B}{\mu_r \mu_0}$.
Substituting the values: $H = \frac{1.0}{400 \times 4\pi \times 10^{-7}} = \frac{1}{1600\pi \times 10^{-7}} = \frac{1}{16\pi \times 10^{-5}}$.
This simplifies to $H = \frac{10^5}{16\pi} = \frac{1}{16\pi} \times 10^5 \text{ A/m}$.
Comparing this with the given expression $\alpha \times 10^5$, we find $\alpha = \frac{1}{16\pi}$.

(A) The magnetic field inside a toroid in vacuum is given by $B_0 = \mu_0 n I$.
When the toroid is filled with a material of magnetic susceptibility $\chi$, the magnetic field becomes $B = \mu_0 (1 + \chi) n I = B_0 (1 + \chi)$.
The increase in the magnetic field is $\Delta B = B - B_0 = B_0 \chi$.
The percentage increase in the magnetic field is given by $\frac{\Delta B}{B_0} \times 100 \% = \chi \times 100 \%$.
Given $\chi = 2.6 \times 10^{-5}$, the percentage increase is $(2.6 \times 10^{-5}) \times 100 \% = 2.6 \times 10^{-3} \%$.

(A) The magnetic field inside a long solenoid is given by $B = \mu_0 n I$, where $n = N/L$ is the number of turns per unit length.
The magnetic flux $\phi$ through a cross-sectional area $A$ is $\phi = B \cdot A = \mu_0 \frac{N}{L} I A$.
The magnetic moment $M$ of a solenoid is given by $M = N I A$.
Substituting $M$ into the flux equation: $\phi = \frac{\mu_0 M}{L}$.
Given $\phi = \frac{\pi}{2} \times 10^{-6} \text{ Wb}$, $L = 0.6 \text{ m}$, and $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$.
$\frac{\pi}{2} \times 10^{-6} = \frac{4\pi \times 10^{-7} \times M}{0.6}$.
$\frac{1}{2} \times 10^{-6} = \frac{4 \times 10^{-7} \times M}{0.6}$.
$0.5 \times 10^{-6} = \frac{4 \times 10^{-7} \times M}{0.6}$.
$M = \frac{0.5 \times 10^{-6} \times 0.6}{4 \times 10^{-7}} = \frac{0.3 \times 10^{-6}}{4 \times 10^{-7}} = \frac{0.3}{0.4} = 0.75 \text{ Am}^2$.

(B) For a long solenoid, the magnetic field induction $(B)$ at its centre is given by the formula: $B = \mu_0 n I$.
By definition, the magnetic field intensity $(H)$ is related to the magnetic field induction $(B)$ in free space by the relation: $H = B / \mu_0$.
Substituting the value of $B$ into the equation for $H$, we get: $H = (\mu_0 n I) / \mu_0$.
Therefore, $H = n I$.
Thus, the correct option is $B$.

(A) The magnetic field $(B)$ inside an ideal solenoid is given by the formula:
$B = \mu n I$
Where:
- $\mu$ is the permeability of the medium inside the solenoid.
- $n$ is the number of turns per unit length.
- $I$ is the current flowing through the solenoid.
From the formula, it is clear that the magnetic induction depends on the permeability of the medium, the number of turns per unit length, and the current flowing through it. It does not depend on the radius of the wire.
Therefore, the correct quantities are $(a)$, $(c)$, and $(d)$.

(C) The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_0 n I$, where $n$ is the number of turns per unit length and $I$ is the current flowing through the solenoid.
Let the initial magnetic field be $B_1 = \mu_0 n I$.
According to the problem, the new current $I' = 3I$ and the new number of turns per unit length $n' = n/2$.
The new magnetic field $B'$ is given by $B' = \mu_0 n' I' = \mu_0 (n/2) (3I) = \frac{3}{2} \mu_0 n I$.
Substituting $B = \mu_0 n I$, we get $B' = \frac{3}{2} B$.

(D) The magnetic intensity $H$ at the centre of a long solenoid is given by the formula $H = nI$, where $n$ is the number of turns per unit length and $I$ is the current flowing through it.
Given:
Total number of turns $N = 1000$
Length of the solenoid $L = 4 \text{ m}$
Current $I = 4 \text{ A}$
First, calculate the number of turns per unit length $n = N / L = 1000 / 4 = 250 \text{ turns/m}$.
Now, substitute the values into the formula:
$H = 250 \times 4 = 1000 \text{ A/m} = 10^{3} \text{ A/m}$.
Therefore, the correct option is $D$.

(C) According to Ampere's Circuital Law, the line integral of the magnetic field $\vec{B}$ around a closed loop is given by $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$.
Here, $I_{\text{enclosed}}$ is the net current passing through the area enclosed by the loop $P$.
The currents inside the loop are $I_1 = 3 \text{ A}$ (upwards) and $I_2 = 1.5 \text{ A}$ (downwards).
Taking the upward direction as positive, the net enclosed current is $I_{\text{enclosed}} = 3 \text{ A} - 1.5 \text{ A} = 1.5 \text{ A}$.
The current outside the loop does not contribute to the line integral.
Therefore, $\oint \vec{B} \cdot d\vec{l} = \mu_0 (1.5 \text{ A}) = 1.5 \mu_0$.