Magnetic Effect of Current
Two insulated circular loop A and B of radius '$a$' carrying a current of '$\mathrm{I}$' in the anti clockwise direction as shown in the figure. The magnitude of the magnetic induction at the centre will be :
Two particles $X$ and $Y$ having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii $R_1$ and $R_2$ respectively. The mass ratio of $X$ and $Y$ is :
A proton moving with a constant velocity passes through a region of space without any change in its velocity. If $\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{B}}$ represent the electric and magnetic fields respectively, then the region of space may have :
(A) $\mathrm{E}=0, \mathrm{~B}=0$
(B) $\mathrm{E}=0, \mathrm{~B} \neq 0$
(C) $\mathrm{E} \neq 0, \mathrm{~B}=0$
(D) $\mathrm{E} \neq 0, \mathrm{~B} \neq 0$
Choose the most appropriate answer from the options given below :
A straight magnetic strip has a magnetic moment of $44 \mathrm{~Am}^2$. If the strip is bent in a semicircular shape, its magnetic moment will be ________ $\mathrm{Am}^2$.
(given $\pi=\frac{22}{7}$)
Explanation:
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 $28 \, \text{Am}^2$.
A square loop of edge length $2 \mathrm{~m}$ carrying current of $2 \mathrm{~A}$ is placed with its edges parallel to the $x$-$y$ axis. A magnetic field is passing through the $x$-$y$ plane and expressed as $\vec{B}=B_0(1+4 x) \hat{k}$, where $B_o=5 T$. The net magnetic force experienced by the loop is _________ $\mathrm{N}$.
Explanation:
Due to constant component of magnetic field $F = 0$
Due to variable component
$\begin{aligned} & F_1=0 \\ & \text { and, } F_2+F_3=0 \\ & \begin{aligned} \text { and, } F_4 & =\left(\mathrm{B}_0 4_x\right) i \mathrm{~L} \\ & =5 \times 4 \times 2 \times 2 \times 2 \\ & =160 \mathrm{~N} \end{aligned} \end{aligned}$
A square loop PQRS having 10 turns, area $3.6 \times 10^{-3} \mathrm{~m}^2$ and resistance $100 \Omega$ is slowly and uniformly being pulled out of a uniform magnetic field of magnitude $\mathrm{B}=0.5 \mathrm{~T}$ as shown. Work done in pulling the loop out of the field in $1.0 \mathrm{~s}$ is _________ $\times 10^{-6} \mathrm{~J}$.
Explanation:
$\begin{aligned} & A = 36 \times 10^{-4} \mathrm{~m}^2 \\ & I= 6 \times 10^{-2} \mathrm{~m} \\ & =6 \mathrm{~cm} \\ V & =\frac{6 \mathrm{~cm}}{1 \mathrm{sec}}=6 \mathrm{~cm} / \mathrm{s} \\ \varepsilon & =B / \mathrm{vn}^2=0.5 \times \frac{6}{100} \times \frac{6}{100} \\ & =18 \times 10^{-4} \mathrm{~V} \\ E & =\frac{n^2 \varepsilon^2}{R} t=100 \times \frac{18 \times 18 \times 10^{-4} \times 10^{-4}}{10^2} \times 1 \\ & =324 \times 10^{-10} \times 10^2 \\ & =3.24 \times 10^{-6} \end{aligned}$
An electron with kinetic energy $5 \mathrm{~eV}$ enters a region of uniform magnetic field of 3 $\mu \mathrm{T}$ perpendicular to its direction. An electric field $\mathrm{E}$ is applied perpendicular to the direction of velocity and magnetic field. The value of E, so that electron moves along the same path, is __________ $\mathrm{NC}^{-1}$.
(Given, mass of electron $=9 \times 10^{-31} \mathrm{~kg}$, electric charge $=1.6 \times 10^{-19} \mathrm{C}$)
Explanation:
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:
$\boxed{4 \, \text{N/C}}$
A coil having 100 turns, area of $5 \times 10^{-3} \mathrm{~m}^2$, carrying current of $1 \mathrm{~mA}$ is placed in uniform magnetic field of $0.20 \mathrm{~T}$ such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through $90^{\circ}$ is _________ $\mu \mathrm{J}$.
Explanation:
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 $100 \mu \mathrm{J}$.
A circular coil having 200 turns, $2.5 \times 10^{-4} \mathrm{~m}^2$ area and carrying $100 \mu \mathrm{A}$ current is placed in a uniform magnetic field of $1 \mathrm{~T}$. Initially the magnetic dipole moment $(\vec{M})$ was directed along $\vec{B}$. Amount of work, required to rotate the coil through $90^{\circ}$ from its initial orientation such that $\vec{M}$ becomes perpendicular to $\vec{B}$, is ________ $\mu$J.
Explanation:
$\begin{aligned} & W=U_f-U_i=(-M B \cos 90)-(-M B \cos 0) \\ & \Rightarrow W=M B=N i A B=5 \mu \mathrm{J} \end{aligned}$
A solenoid of length $0.5 \mathrm{~m}$ has a radius of $1 \mathrm{~cm}$ and is made up of '$\mathrm{m}$' number of turns. It carries a current of $5 \mathrm{~A}$. If the magnitude of the magnetic field inside the solenoid is $6.28 \times 10^{-3} \mathrm{~T}$ then the value of $\mathrm{m}$ is __________.
Explanation:
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.
A 2A current carrying straight metal wire of resistance $1 \Omega$, resistivity $2 \times 10^{-6} \Omega \mathrm{m}$, area of cross-section $10 \mathrm{~mm}^2$ and mass $500 \mathrm{~g}$ is suspended horizontally in mid air by applying a uniform magnetic field $\vec{B}$. The magnitude of B is ________ $\times 10^{-1} \mathrm{~T}$ (given, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$).
Explanation:
$\begin{aligned} & i L B=m g \text { and } L=\frac{A R}{\rho} \\ & \begin{aligned} \therefore B & =\frac{m g \rho}{i A R} \\ & =\frac{0.5 \times 10 \times 2 \times 10^{-6}}{2 \times 10 \times 10^{-6} \times 1} \\ & =0.5 \mathrm{~T} \end{aligned} \end{aligned}$
Two parallel long current carrying wire separated by a distance $2 r$ are shown in the figure. The ratio of magnetic field at $A$ to the magnetic field produced at $C$ is $\frac{x}{7}$. The value of $x$ is __________.
Explanation:
At point $A$
$B_A=\frac{\mu_0 I}{2 \pi r}+\frac{\mu_0(2 I)}{2 \pi(3 r)}$
At point $C$
$\begin{aligned} & B_C=\frac{\mu_0 I}{2 \pi(3 r)}+\frac{\mu_0(2 I)}{r} \\ & \Rightarrow \frac{B_A}{B_C}=\frac{5}{7} \end{aligned}$
A rod of length $60 \mathrm{~cm}$ rotates with a uniform angular velocity $20 \mathrm{~rad} \mathrm{s}^{-1}$ about its perpendicular bisector, in a uniform magnetic filed $0.5 T$. The direction of magnetic field is parallel to the axis of rotation. The potential difference between the two ends of the rod is _________ V.
Explanation:
Both end having same potential, so potential difference between end will be zero.
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:
$\begin{aligned} & \mathrm{B}=\left(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{a}}\right) \times 2=\frac{4 \pi \times 10^{-7} \times 10}{\pi \times\left(\frac{5}{2} \times 10^{-2}\right)} \\ & =16 \times 10^{-5}=160 \mu \mathrm{T} \end{aligned}$
An electron is moving along the positive $\mathrm{x}$-axis. If the uniform magnetic field is applied parallel to the negative z-axis, then
A. The electron will experience magnetic force along positive y-axis
B. The electron will experience magnetic force along negative y-axis
C. The electron will not experience any force in magnetic field
D. The electron will continue to move along the positive $\mathrm{x}$-axis
E. The electron will move along circular path in magnetic field
Choose the correct answer from the options given below:
The source of time varying magnetic field may be
(A) a permanent magnet
(B) an electric field changing linearly with time
(C) direct current
(D) a decelerating charge particle
(E) an antenna fed with a digital signal
Choose the correct answer from the options given below:
An electron is allowed to move with constant velocity along the axis of current carrying straight solenoid.
A. The electron will experience magnetic force along the axis of the solenoid.
B. The electron will not experience magnetic force.
C. The electron will continue to move along the axis of the solenoid.
D. The electron will be accelerated along the axis of the solenoid.
E. The electron will follow parabolic path-inside the solenoid.
Choose the correct answer from the options given below:
A charge particle moving in magnetic field B, has the components of velocity along B as well as perpendicular to B. The path of the charge particle will be
A long straight wire of circular cross-section (radius a) is carrying steady current I. The current I is uniformly distributed across this cross-section. The magnetic field is
As shown in the figure, a long straight conductor with semicircular arc of radius $\frac{\pi}{10}$m is carrying current $\mathrm{I=3A}$. The magnitude of the magnetic field, at the center O of the arc is :
(The permeability of the vacuum $=4\pi\times10^{-7}~\mathrm{NA}^{-2}$)
Find the magnetic field at the point $\mathrm{P}$ in figure. The curved portion is a semicircle connected to two long straight wires.
A rod with circular cross-section area $2 \mathrm{~cm}^{2}$ and length $40 \mathrm{~cm}$ is wound uniformly with 400 turns of an insulated wire. If a current of $0.4 \mathrm{~A}$ flows in the wire windings, the total magnetic flux produced inside windings is $4 \pi \times 10^{-6} \mathrm{~Wb}$. The relative permeability of the rod is
(Given : Permeability of vacuum $\mu_{0}=4 \pi \times 10^{-7} \mathrm{NA}^{-2}$)
As shown in the figure, a current of $2 \mathrm{~A}$ flowing in an equilateral triangle of side $4 \sqrt{3} \mathrm{~cm}$. The magnetic field at the centroid $\mathrm{O}$ of the triangle is
(Neglect the effect of earth's magnetic field)
A current carrying rectangular loop PQRS is made of uniform wire. The length $P R=Q S=5 \mathrm{~cm}$ and $P Q=R S=100 \mathrm{~cm}$. If ammeter current reading changes from I to $2 I$, the ratio of magnetic forces per unit length on the wire $P Q$ due to wire $R S$ in the two cases respectively $\left(f_{P Q}^I: f_{P Q}^{2 t}\right)$ is:
A massless square loop, of wire of resistance $10 \Omega$, supporting a mass of $1 \mathrm{~g}$, hangs vertically with one of its sides in a uniform magnetic field of $10^{3} \mathrm{G}$, directed outwards in the shaded region. A dc voltage $\mathrm{V}$ is applied to the loop. For what value of $\mathrm{V}$, the magnetic force will exactly balance the weight of the supporting mass of $1 \mathrm{~g}$ ?
(If sides of the loop $=10 \mathrm{~cm}, \mathrm{~g}=10 \mathrm{~ms}^{-2}$)
The magnetic moments associated with two closely wound circular coils $\mathrm{A}$ and $\mathrm{B}$ of radius $\mathrm{r}_{\mathrm{A}}=10$ $\mathrm{cm}$ and $\mathrm{r}_{\mathrm{B}}=20 \mathrm{~cm}$ respectively are equal if : (Where $\mathrm{N}_{\mathrm{A}}, \mathrm{I}_{\mathrm{A}}$ and $\mathrm{N}_{\mathrm{B}}, \mathrm{I}_{\mathrm{B}}$ are number of turn and current of $\mathrm{A}$ and $\mathrm{B}$ respectively)
The electric current in a circular coil of four turns produces a magnetic induction 32 T at its centre. The coil is unwound and is rewound into a circular coil of single turn, the magnetic induction at the centre of the coil by the same current will be :
The magnitude of magnetic induction at mid point $\mathrm{O}$ due to current arrangement as shown in Fig will be
A single current carrying loop of wire carrying current I flowing in anticlockwise direction seen from +ve $\mathrm{z}$ direction and lying in $x y$ plane is shown in figure. The plot of $\hat{j}$ component of magnetic field (By) at a distance '$a$' (less than radius of the coil) and on $y z$ plane vs $z$ coordinate looks like
For a moving coil galvanometer, the deflection in the coil is 0.05 rad when a current of 10 mA is passes through it. If the torsional constant of suspension wire is $4.0\times10^{-5}\mathrm{N~m~rad^{-1}}$, the magnetic field is 0.01T and the number of turns in the coil is 200, the area of each turn (in cm$^2$) is :
Match List I with List II
| List I (Current configuration) |
List II (Magnitude of Magnetic Field at point O) |
||
|---|---|---|---|
| A. |  |
I. | ${B_0} = {{{\mu _0}I} \over {4\pi r}}[\pi + 2]$ |
| B. |  |
II. | ${B_0} = {{{\mu _0}} \over {4 }}{I \over r}$ |
| C. |  |
III. | ${B_0} = {{{\mu _0}I} \over {2\pi r}}[\pi - 1]$ |
| D. |  |
IV. | ${B_0} = {{{\mu _0}I} \over {4\pi r}}[\pi + 1]$ |
Choose the correct answer from the options given below :
A long solenoid is formed by winding 70 turns cm$^{-1}$. If 2.0 A current flows, then the magnetic field produced inside the solenoid is ____________ ($\mu_0=4\pi\times10^{-7}$ TmA$^{-1}$)
Two long straight wires P and Q carrying equal current 10A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F$_1$. If distance between wires is halved and currents on them are doubled, force F$_2$ on 10 cm length of wire P will be:
A circular loop of radius $r$ is carrying current I A. The ratio of magnetic field at the center of circular loop and at a distance r from the center of the loop on its axis is :
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} $