The magnetic field existing in a region is given by $\vec{B}=0.2(1+2 x) \hat{k}$. A square loop of edge $50 \mathrm{~cm}$ carrying 0.5 A current is placed in $x$-$y$ plane with its edges parallel to the $x$-$y$ axes, as shown in figure. The magnitude of the net magnetic force experienced by the loop is _________ $\mathrm{mN}$.
Explanation:
$\begin{aligned} & \vec{F}_{B C}+\vec{F}_{D A}=0 \\ & \vec{F}_{A B}=i l B=0.5 \times 0.5(5)=1.25 \mathrm{~N} \times 0.2=0.25 \mathrm{~N} \\ & \vec{F}_{C D}=0.5 \times 0.5(6)=1.5 \times 0.2=0.3 \mathrm{~N} \\ & F_{\text {net }}=0.05 \mathrm{~N} \\ & \quad=50 \mathrm{mN} \end{aligned}$
Explanation:
If an electric current of $4 \pi \sqrt{3}$ A is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be $x \times 10^{-7} \mathrm{~T}$.
The value of $x$ is _________.
Explanation:
$\begin{aligned} & B=6\left(\frac{\mu_0 I}{4 \pi r}\right)\left(\sin 30^{\circ}+\sin 30^{\circ}\right) \\\\ & =6 \frac{10^{-7} \times 4 \pi \sqrt{3}}{\left(\frac{\sqrt{3} \times 4 \pi}{2 \times 6}\right)} \\\\ & =72 \times 10^{-7} \mathrm{~T}\end{aligned}$
Two circular coils $P$ and $Q$ of 100 turns each have same radius of $\pi \mathrm{~cm}$. The currents in $P$ and $R$ are $1 A$ and $2 A$ respectively. $P$ and $Q$ are placed with their planes mutually perpendicular with their centers coincide. The resultant magnetic field induction at the center of the coils is $\sqrt{x} ~m T$, where $x=$ __________.
[Use $\mu_0=4 \pi \times 10^{-7} \mathrm{~TmA}^{-1}$]
Explanation:
$\begin{aligned} & \mathrm{B}_{\mathrm{P}}=\frac{\mu_0 \mathrm{Ni}_1}{2 \mathrm{r}}=\frac{\mu_0 \times 1 \times 100}{2 \pi}=2 \times 10^{-3} \mathrm{~T} \\ & \mathrm{~B}_{\mathrm{Q}}=\frac{\mu_0 \mathrm{Ni}_2}{2 \mathrm{r}}=\frac{\mu_0 \times 2 \times 100}{2 \pi}=4 \times 10^{-3} \mathrm{~T} \\ & \mathrm{~B}_{\text {net }}=\sqrt{\mathrm{B}_{\mathrm{P}}^2+\mathrm{B}_{\mathrm{Q}}^2} \\ & =\sqrt{20} \mathrm{mT} \\ & \mathrm{x}=20 \end{aligned}$
An electron moves through a uniform magnetic field $\vec{B}=B_0 \hat{i}+2 B_0 \hat{j} T$. At a particular instant of time, the velocity of electron is $\vec{u}=3 \hat{i}+5 \hat{j} \mathrm{~m} / \mathrm{s}$. If the magnetic force acting on electron is $\vec{F}=5 e \hat{k} N$, where $e$ is the charge of electron, then the value of $B_0$ is _________ $T$.
Explanation:
$\begin{aligned} & \overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \\ & 5 \mathrm{e} \hat{\mathrm{k}}=\mathrm{e}(3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}) \times\left(\mathrm{B}_0 \hat{\mathrm{i}}+2 \mathrm{~B}_0 \hat{\mathrm{j}}\right) \\ & 5 \mathrm{e} \hat{\mathrm{k}}=\mathrm{e}\left(6 \mathrm{~B}_0 \hat{\mathrm{k}}-5 \mathrm{~B}_0 \hat{\mathrm{k}}\right) \\ & \Rightarrow \mathrm{B}_0=5 \mathrm{~T} \end{aligned}$
The current of $5 \mathrm{~A}$ flows in a square loop of sides $1 \mathrm{~m}$ is placed in air. The magnetic field at the centre of the loop is $X \sqrt{2} \times 10^{-7} T$. The value of $X$ is _________.
Explanation:
$\begin{aligned} & \mathrm{B}=4 \times \frac{\mu_0 \mathrm{i}}{4 \pi(1 / 2)}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right) \\ & =4 \times 10^{-7} \times 5 \times 2 \times \sqrt{2} \\ & -40 \sqrt{2} \times 10^{-7} \mathrm{~T} \end{aligned}$
A charge of $4.0 \mu \mathrm{C}$ is moving with a velocity of $4.0 \times 10^6 \mathrm{~ms}^{-1}$ along the positive $y$ axis under a magnetic field $\vec{B}$ of strength $(2 \hat{k}) \mathrm{T}$. The force acting on the charge is $x \hat{i} N$. The value of $x$ is __________.
Explanation:
$\begin{aligned} \mathrm{q} & =4 \mu \mathrm{C}, \overrightarrow{\mathrm{v}}=4 \times 10^6 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s} \\ \overrightarrow{\mathrm{B}} & =2 \hat{\mathrm{k} T} \\ \overrightarrow{\mathrm{F}} & =\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \\ & =4 \times 10^{-6}\left(4 \times 10^6 \hat{\mathrm{j}} \times 2 \hat{\mathrm{k}}\right) \\ & =4 \times 10^{-6} \times 8 \times 10^6 \hat{\mathrm{i}} \\ \overrightarrow{\mathrm{F}} & =32 \hat{\mathrm{i}} \mathrm{N} \\ \mathrm{x} & =32 \end{aligned}$
The magnetic field at the centre of a wire loop formed by two semicircular wires of radii $R_1=2 \pi \mathrm{m}$ and $R_2=4 \pi \mathrm{m}$, carrying current $\mathrm{I}=4 \mathrm{~A}$ as per figure given below is $\alpha \times 10^{-7} \mathrm{~T}$. The value of $\alpha$ is ________. (Centre $\mathrm{O}$ is common for all segments)
Explanation:
$\begin{aligned} & \frac{\mu_0 \mathrm{i}}{2 \mathrm{R}_2}\left(\frac{\pi}{2 \pi}\right) \otimes+\frac{\mu_0 \mathrm{i}}{2 \mathrm{R}_1}\left(\frac{\pi}{2 \pi}\right) \otimes \\ & \left(\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}_2}+\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}_1}\right) \otimes \\ & \frac{4 \pi \times 10^{-7} \times 4}{4 \times 4 \pi}+\frac{4 \pi \times 10^{-7} \times 4}{4 \times 2 \pi} \\ & =3 \times 10^{-7}=\alpha \times 10^{-7} \\ & \alpha=3 \end{aligned}$
Two long, straight wires carry equal currents in opposite directions as shown in figure. The separation between the wires is $5.0 \mathrm{~cm}$. The magnitude of the magnetic field at a point $\mathrm{P}$ midway between the wires is _______ $\mu \mathrm{T}$
(Given : $\mu_0=4 \pi \times 10^{-7} \mathrm{TmA}^{-1}$)
Explanation:
$\begin{aligned} & \mathrm{B}=\left(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{a}}\right) \times 2=\frac{4 \pi \times 10^{-7} \times 10}{\pi \times\left(\frac{5}{2} \times 10^{-2}\right)} \\ & =16 \times 10^{-5}=160 \mu \mathrm{T} \end{aligned}$
Explanation:
$\mathbf{B}=\frac{\mu_0}{4 \pi} \frac{q v \sin \theta}{r^2}$
where $\mu_0$ is the permeability of free space, $q$ is the charge of the moving particle, $v$ is the speed of the particle, $\theta$ is the angle between the velocity vector and the position vector from the particle to the point where we want to calculate the magnetic field, and $r$ is the distance between the particle and the point where we want to calculate the magnetic field.
In this case, we're interested in the magnetic field produced by the electron moving in a circular orbit around the nucleus of a hydrogen atom. Since the orbit is circular, the angle between the velocity vector and the position vector is 90 degrees, so $\sin \theta = 1$. We can substitute the known values into the formula to find the magnetic field:
$\mathbf{B}=\frac{\mu_0}{4 \pi} \frac{q v \sin \theta}{r^2} = \frac{\mu_0}{4 \pi} \frac{e v}{r^2}$
where $e$ is the charge of an electron. We know that the radius of the orbit is $0.52 \mathrm{~A}^{\circ}$, which is equivalent to $0.52 \times 10^{-10} \mathrm{m}$.
Substituting the values, we get:
$\mathbf{B}=\frac{\mu_0}{4 \pi} \frac{e v}{r^2} =\frac{10^{-7} \times 1.6 \times 10^{-19} \times 6.76 \times 10^6}{0.52 \times 0.52 \times 10^{-20}} = 40 ~\mathrm{T}$
This means that the magnetic field produced by the electron moving in a circular orbit around the nucleus of a hydrogen atom is 40 tesla, which is an incredibly strong magnetic field.
A straight wire $\mathrm{AB}$ of mass $40 \mathrm{~g}$ and length $50 \mathrm{~cm}$ is suspended by a pair of flexible leads in uniform magnetic field of magnitude $0.40 \mathrm{~T}$ as shown in the figure. The magnitude of the current required in the wire to remove the tension in the supporting leads is ___________ A.
$\left(\right.$ Take $g=10 \mathrm{~ms}^{-2}$ ).
Explanation:
For equilibrium, the magnetic force on the wire should balance the weight of the wire. Therefore, we can write:
$ \mathrm{Mg}=\mathrm{I} \ell \mathrm{B} $
where $\mathrm{M}$ is the magnetic force on the wire, $\mathrm{g}$ is the acceleration due to gravity, $\mathrm{I}$ is the current flowing through the wire, $\ell$ is the length of the wire, and $\mathrm{B}$ is the magnitude of the magnetic field.
Solving for $\mathrm{I}$, we get:
$ \mathrm{I}=\frac{\mathrm{mg}}{\ell \mathrm{B}} $
Substituting the given values, we get:
$ \mathrm{I}=\frac{40 \times 10^{-3} \times 10}{50 \times 10^{-2} \times 0.4}=2 \mathrm{~A} $
Therefore, the magnitude of the current required in the wire to remove the tension in the supporting leads is 2 A.
A straight wire carrying a current of $14 \mathrm{~A}$ is bent into a semi-circular arc of radius $2.2 \mathrm{~cm}$ as shown in the figure. The magnetic field produced by the current at the centre $(\mathrm{O})$ of the arc. is ____________ $\times ~10^{-4} \mathrm{~T}$
Explanation:
The ratio of magnetic field at the centre of a current carrying coil of radius $r$ to the magnetic field at distance $r$ from the centre of coil on its axis is $\sqrt{x}: 1$. The value of $x$ is __________
Explanation:
The magnetic field at the center of a loop (B1) is given by
$ B_1 = \frac{\mu_0 I}{2r} $
The magnetic field on the axis of the loop at a distance ( r ) from the center (B2) is given by
$ B_2 = \frac{\mu_0 Ir^2}{2(r^2 + d^2)^{3/2}} $
where ( d ) is the distance from the center of the coil along the axis. Since ( d = r ), we get
$ B_2 = \frac{\mu_0 I}{4\sqrt{2}r} $
The ratio of $ B_1 $ to $ B_2 $ is
$ \frac{B_1}{B_2} = \frac{\mu_0 I}{2r} \times \frac{4\sqrt{2}r}{\mu_0 I} = \sqrt{8} : 1 $
So, the value of ( x ) is 8.
A proton with a kinetic energy of $2.0 ~\mathrm{eV}$ moves into a region of uniform magnetic field of magnitude $\frac{\pi}{2} \times 10^{-3} \mathrm{~T}$. The angle between the direction of magnetic field and velocity of proton is $60^{\circ}$. The pitch of the helical path taken by the proton is __________ $\mathrm{cm}$. (Take, mass of proton $=1.6 \times 10^{-27} \mathrm{~kg}$ and Charge on proton $=1.6 \times 10^{-19} \mathrm{C}$ ).
Explanation:
Given a proton with a kinetic energy of 2 eV, moving into a region of uniform magnetic field of magnitude $\frac{\pi}{2} \times 10^{-3} T$, and with an angle of $60^{\circ}$ between the direction of the magnetic field and the velocity of the proton, we want to determine the pitch of the helical path taken by the proton.
- First, calculate the proton's speed (v) using the kinetic energy (K.E) formula:
$v = \sqrt{\frac{2 \times KE}{m}}$
- Next, find the component of the velocity in the direction of the magnetic field (parallel component):
$v_{\parallel} = v \cos \theta$
In this case, θ is given as $60^{\circ}$, so $\cos \theta = \frac{1}{2}$.
- The pitch of a charged particle moving in a magnetic field with an angle θ to the direction of the magnetic field is given by the formula:
$p = \frac{2 \pi m v_{\parallel}}{qB}$
- Substitute the values for the mass of the proton (m), the kinetic energy (KE), the charge of the proton (q), and the magnetic field (B) into the formula:
$p = \frac{2 \pi \times \sqrt{2mKE} \times \frac{1}{2} \times 2}{qB}$
- After substituting the given values, the pitch of the helical path is found to be:
$p = 0.4 \, m = 40 \, cm$
In conclusion, the pitch of the helical path taken by the proton in the magnetic field is 40 cm.
Two identical circular wires of radius $20 \mathrm{~cm}$ and carrying current $\sqrt{2} \mathrm{~A}$ are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wires is __________ $\times 10^{-8} \mathrm{~T}$.
(Take $\pi=3.14$)
Explanation:
$\begin{aligned} \mathbf{B}_{\text {net }} & =\frac{\mu_0 i}{2 r} \hat{\mathbf{i}}+\frac{\mu_0 i}{2 r} \hat{\mathbf{j}}=\frac{\mu_0 i}{2 r} \hat{\mathbf{k}} \sqrt{2} \\\\ & =4 \pi \times 10^{-7} \times \sqrt{2} \times \sqrt{2} \times \frac{1}{2 \times 0.2} \\\\ & =2 \times 3.14 \times 10^{-6}=628 \times 10^{-8} \mathrm{~T}\end{aligned}$
A charge particle of $2 ~\mu \mathrm{C}$ accelerated by a potential difference of $100 \mathrm{~V}$ enters a region of uniform magnetic field of magnitude $4 ~\mathrm{mT}$ at right angle to the direction of field. The charge particle completes semicircle of radius $3 \mathrm{~cm}$ inside magnetic field. The mass of the charge particle is __________ $\times 10^{-18} \mathrm{~kg}$
Explanation:
$m=\frac{r^2 q^2 B^2}{2 k}$
$ \begin{aligned} \mathrm{m}= & \frac{\frac{1}{100} \times \frac{3}{100} \times 2 \times 2 \times 4 \times 10^{-3} \times 4 \times 10^{-3} \times 10^{-12}}{2 \times(100) \times 10^{-6}} \\\\ & =144 \times 10^{-18} \mathrm{~kg} \end{aligned} $
Two long parallel wires carrying currents 8A and 15A in opposite directions are placed at a distance of 7 cm from each other. A point P is at equidistant from both the wires such that the lines joining the point P to the wires are perpendicular to each other. The magnitude of magnetic field at P is _____________ $\times~10^{-6}$ T.
(Given : $\sqrt2=1.4$)
Explanation:
$ \begin{aligned} & B_1=\frac{\mu_0 i_1}{2 \pi d} \quad B_2=\frac{\mu_0 i_2}{2 \pi d} \\\\ & B_{\text {net }}=\sqrt{B_1^2+B_2^2} = \frac{\mu_0}{2 \pi d} \sqrt{i_1^2+i_2^2} \\\\ & = \frac{4 \pi \times 10^{-7}}{2 \pi \times(7 / \sqrt{2}) \times 10^{-2}} \times \sqrt{15^2+8^2}\left(\text {As }d=\frac{7}{\sqrt{2}} \mathrm{~cm}\right) \\\\ & = 68 \times 10^{-6} \mathrm{~T} \end{aligned} $
A single turn current loop in the shape of a right angle triangle with sides 5 cm, 12 cm, 13 cm is carrying a current of 2 A. The loop is in a uniform magnetic field of magnitude 0.75 T whose direction is parallel to the current in the 13 cm side of the loop. The magnitude of the magnetic force on the 5 cm side will be $\frac{x}{130}$ N. The value of $x$ is ____________.
Explanation:
$ \begin{aligned} & =2 \times \frac{5}{100} \times 0.75 \times \frac{12}{13} \\\\ & =\frac{9}{130} \\\\ & \therefore \quad x=9 \end{aligned} $
A closely wounded circular coil of radius 5 cm produces a magnetic field of $37.68 \times 10^{-4} \mathrm{~T}$ at its center. The current through the coil is _________A.
[Given, number of turns in the coil is 100 and $\pi=3.14$]
Explanation:
$B = {{{\mu _0}nI} \over {2R}}$
$37.68 \times {10^{ - 4}} = {{4\pi \times {{10}^{ - 7}}100\,I} \over {2 \times 5 \times {{10}^{ - 2}}}}$
$I = {{300\,A} \over {100}}$
$ = 3\,A$
A wire of length $314 \mathrm{~cm}$ carrying current of $14 \mathrm{~A}$ is bent to form a circle. The magnetic moment of the coil is ________ A $-\mathrm{m}^{2}$. [Given $\pi=3.14$]
Explanation:
$R = {l \over {2\pi }} = {{314} \over {2 \times 3.14}} = 50$ cm
$\mu = \pi {R^2}i$
$ = 14 \times 3.14 \times {(0.5)^2}$
$ = 11$ A-m2
A singly ionized magnesium atom (A = 24) ion is accelerated to kinetic energy 5 keV, and is projected perpendicularly into a magnetic field B of the magnitude 0.5 T. The radius of path formed will be _____________ cm.
Explanation:
To calculate the radius of the path formed by a singly ionized magnesium atom in a magnetic field, the following formula is used:
$ R = \frac{mv}{qB} $
This can be rewritten using kinetic energy (KE):
$ R = \frac{\sqrt{2m \cdot KE}}{qB} $
Here's how we calculate it for a magnesium ion:
Atomic Mass (A): 24
Mass of a Nucleon: $1.67 \times 10^{-27} $ kg (approximate mass of a proton or neutron)
Charge (q): $1.6 \times 10^{-19} $ C (since the magnesium ion is singly ionized)
Magnetic Field (B): 0.5 T
Kinetic Energy (KE): 5 keV = $5 \times 1.6 \times 10^{-16} $ J
Plug in these values to find the radius:
$ R = \frac{\sqrt{2 \times 24 \times 1.67 \times 10^{-27} \times 5 \times 1.6 \times 10^{-16}}}{1.6 \times 10^{-19} \times 0.5} $
This simplifies to:
$ R = 10.009 \, \text{cm} \approx 10 \, \text{cm} $
Thus, the radius of the path is approximately 10 cm.
A deuteron and a proton moving with equal kinetic energy enter into a uniform magnetic field at right angle to the field. If rd and rp are the radii of their circular paths respectively, then the ratio ${{{r_d}} \over {{r_p}}}$ will be $\sqrt{x}$ : 1 where x is __________.
Explanation:
$R = {{\sqrt {2mK} } \over {qB}}$
So, ${{{r_d}} \over {{r_p}}} = {{\sqrt {{m_d}} /{q_d}} \over {\sqrt {{m_p}} /{q_p}}}$
$ = \sqrt 2 $
So $x = 2$
Two 10 cm long, straight wires, each carrying a current of 5A are kept parallel to each other. If each wire experienced a force of 10$-$5 N, then separation between the wires is ____________ cm.
Explanation:
${{dF} \over {dl}} = {{{\mu _0}{i_1}{i_2}} \over {2\pi d}}$
So ${{2 \times {{10}^{ - 7}} \times 5 \times 5} \over d} = {{{{10}^{ - 5}}} \over {10 \times {{10}^{ - 2}}}}$
$d = {{2 \times {{10}^{ - 7}} \times 5 \times 5} \over {{{10}^{ - 4}}}}$
= 50 mm
= 5 cm
Explanation:
$ \Rightarrow $ ${1 \over 2} = {x \over {499}} \Rightarrow x \simeq 250$
Explanation:
In square ${N_s} = {{24a} \over {4a}} = 6$
${{{M_t}} \over {{M_3}}} = {{{N_t}I{A_t}} \over {{N_s}I{A_s}}}$ [I will be same in both]
$ = {{8 \times {{\sqrt 3 } \over 4} \times {a^2}} \over {6 \times {a^2}}}$
${{{M_t}} \over {{M_s}}} = {1 \over {\sqrt 3 }}$
y = 3
Explanation:
$\overrightarrow \tau = \overrightarrow M \times \overrightarrow B = MB\sin 90^\circ $
$ = MB = {{i\sqrt 3 {l^2}} \over 4}B$
$ = \sqrt 3 \times {10^{ - 5}}$ N $-$ m
Explanation:
$\tau = {M_2}{B_1}\sin 90^\circ $
$ = 1 \times {{{\mu _0}} \over {4\pi }}{{{M_1}} \over {{{(1)}^3}}}1$
= 10$-$7 N.m
Explanation:
A = 3 $ \times $ 10–4 m2
$\tau $ = 1.5 Nm
i = 0.5 A
We know, $\tau = BINA\,sin\theta $
$1.5 = B \times 0.5 \times 500 \times 3 \times {10^{ - 4}}$
$B = {{10000} \over {500}} = 20$ Tesla