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 $\alpha = \frac{1}{16\pi}$.