datavorous/entrance-exam-dataset huggingface dataset:
The magnetic field of an electromagnetic wave is given by:
B→=1.6×10-6cos⁡ 2×107z+6×1015t 2i^+j^Wbm2
The associated electric field will be:

Options:
A E→=4.8×102 cos 2×107z-6×1015t -2 j^+i^Vm
B E→=4.8×102 cos 2×107z+6×1015t i^-2 j^Vm
C E→=4.8×102 cos 2×107z+6×1015t -i^+2j^Vm
D E→=4.8×102 cos 2×107z-6×1015t 2i^+ j^Vm

The correct answer is:
E→=4.8×102 cos 2×107z+6×1015t -i^+2j^Vm

Solution:
Given magnetic field, B→=1.6×10-6cos2×107z+6×1015t2i^+j^Wbm2
The relation between magnetic field and electric field is E=cB=(3×108)×(1.6×10-6×5)=4.8×102×5
The direction of light propagation will be along E→×B→ .
E→=4.8×102cos2×107z+6×1015t-i^+2j^Vm

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The magnetic field of earth at the equator is approximately 4 × 1 0 - 5 T . The radius of earth is 6.4 × 1 0 6 m . Then the dipole moment of the earth will be nearly of the order of :

Options:
A 1020 A m2
B 1016 A m2
C 1010 A m2
D 1023 A m2

The correct answer is:
1023 A m2

Solution:
B = μ 0 4 π · M d 3 = 1 0 - 7 × M 6.4 × 1 0 6 3
∴ 4 × 1 0 - 5 = 1 0 - 7 × M 6.4 × 1 0 6 3
∴ M = 6.4 × 1 0 6 3 × 4 × 1 0 - 5 1 0 - 7
= 6.4 3 × 1 0 1 8 × 4 × 1 0 2
= 6.4 3 × 4 × 1 0 2 0
= 1 . 0 4 8 × 1 0 2 3 ≈ 10 23 A m 2

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2014 (09 Apr Online)
The magnetic field of earth at the equator is approximately \( 4 \times 10^{-5} \mathrm{~T} \). The radius of earth is \( 6.4 \times 10^{6} \mathrm{~m} \). Then the dipole moment of the earth will be nearly of the order of :

Options:
A \( 10^{20} \mathrm{Am}^{2} \)
B \( 10^{16} \mathrm{Am}^{2} \)
C \( 10^{10} \mathrm{Am}^{2} \)
D \( 10^{23} \mathrm{Am}^{2} \)

The correct answer is:
\( 10^{23} \mathrm{Am}^{2} \)

Solution:
B = μ 0 4 π · M d 3 = 1 0 - 7 × M 6.4 × 1 0 6 3
∴ 4 × 1 0 - 5 = 1 0 - 7 × M 6.4 × 1 0 6 3
∴ M = 6.4 × 1 0 6 3 × 4 × 1 0 - 5 1 0 - 7
= 6.4 3 × 1 0 1 8 × 4 × 1 0 2
= 6.4 3 × 4 × 1 0 2 0
= 1 . 0 4 8 × 1 0 2 3 ≈ 10 23 A m 2

Topic: Physics, Magnetic Properties of Matter, JEE Main
The magnetic field vector of an electromagnetic wave is given by B=B0i^+j^2coskz-ωt where i^, j^ represents unit vector along x and y-axis respectively. At t=0 s, two electric charges q1 of 4π coulomb and q2 of 2π coulomb located at 0, 0, πk and 0, 0, 3πk, respectively, have the same velocity of 0.5ci^, (where c is the velocity of light ). The ratio of the force acting on charge q1 to q2 is :

Options:
A 1 : 2
B 22 : 1
C 2 : 1
D 2 : 1

The correct answer is:
2 : 1

Solution:
At t=0
B at 0, 0, πk=B0i^+j^2cosπ
B at 0, 0, 3πk=B0i^+j^2cos3π
Force on charged particle q1
F1=q1V→1×B→1
=4π0.5ci^×-B0i^+j^2=4πB0c22-k^
Force on charged particle q2
F2=q2V→2×B→2
=2π0.5ci^×-B0i^+j^2=2πB0c22-k^⇒F1F2=2

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2021 (31 Aug Shift 2)
The magnetic field within a long, straight solenoid with a circular cross-section of radius r is (as shown) increasing at a rate of Î±.

The rate of change of flux through a circle with radius a inside the solenoid, normal to the axis of the solenoid and with centre on the solenoid axis is

Options:
A 2πa2α
B 12Ïa2Î±
C Ïa2Î±
D 32ÏaÎ±

The correct answer is:
πa2α

Solution:
Rate of change of flux through circle with radius a
dϕmdt=AdBdt=πa2α

Topic: Physics, Electromagnetic Induction, JEE Main
The magnetic flux ϕ (in weber) linked with a closed circuit of resistance 8 Ω varies with time (in seconds) as ϕ=5t2−36t+1. The induced current in the circuit at t =2 s is _______ A.

Options:

The correct answer is:
2

Solution:
Induced EMF is given by, ε=−dϕdt=10t−36

Now, it is given that at t=2 s, ε=16 V.

Therefore, i=εR=168=2 A

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2024 (31 Jan Shift 2)
The magnetic flux linked to a circular coil of radius $R$ is:
$\phi=2 t^3+4 t^2+2 t+5 \mathrm{~Wb}$
The magnitude of induced emf in the coil at $t=5 \mathrm{~s}$ is:

Options:
A $108 \mathrm{~V}$
B $197 \mathrm{~V}$
C $150 \mathrm{~V}$
D $192 \mathrm{~V}$

The correct answer is:
$192 \mathrm{~V}$

Solution:
Given:
$\phi=2 t^3+4 t^2+2 t+5,$
& $t=5 \mathrm{~s}$
By Faraday's law of EMI, we have,
$\mathrm{emf}=-\frac{d \phi}{d t}$
(where, the symbols have their usual meanings)
$\begin{aligned}
e=\left|\frac{d \phi}{d t}\right| & =\left|\frac{d}{d t}\left(2 t^3+4 t^2+2 t+5\right)\right| \\
& =6 t^2+8 t+2
\end{aligned}$
$\begin{aligned}
ext{At } t & =5 s \\
e & =6(5)^2+8(5)+2 \\
& =192 \mathrm{~V}
\end{aligned}$

Topic: Physics, Electromagnetic Induction, JEE Main
The magnetic flux linked with a coil at any instant ' $t$ ' is given by $\varphi=$ $5 t^3-100 t+300$, the e.m.f. induced in the coil at $t=2$ second is

Options:
A $-40 \mathrm{~V}$
B $40 \mathrm{~V}$
C $140 \mathrm{~V}$
D $300 \mathrm{~V}$

The correct answer is:
$40 \mathrm{~V}$

Solution:
$e=-\frac{d \phi}{d t}=-\frac{d}{d t}\left(5 t^3-100 t+300\right)$
$=-\left(15 t^2-100\right)$ at $t=2 \mathrm{sec} ; e=40 \mathrm{~V}$

Topic: Physics, Electromagnetic Induction, JEE Main
The magnetic flux linked with a coil (in Wb) is given by the equation ϕ=5t2+3t+16. The magnitude of induced emf in the coil at the fourth second will be:

Options:
A 33 V
B 43 V
C 108 V
D 10 V

The correct answer is:
43 V

Solution:
Given magnetic flux in terms of time, ϕ=5t2+3t+60, and we know that from Faraday's law of electromagnetic induction, rate of change of flux is directly related to induced EMF of the coil,

⇒|ε|=dϕdt=10t+3

At t=4 s, |ε|=40+3=43 V

Topic: Physics, Electromagnetic Induction, JEE Main
The magnetic flux linked with a coil, in webers, is given by the equations $\phi=3 t^2+4 t+9$. Then the magnitude of induced e.m.f. at $t=2$ second will be

Options:
A $2$ volt
B $4$ volt
C $8$ volt
D $16$ volt

The correct answer is:
$16$ volt

Solution:
$e=-\frac{d \phi}{d t}=-\frac{d}{d t}\left(3 t^2+4 t+9\right)=-(6 t+4)$
$e=-[6(2)+4]=-16 \Rightarrow|e|=16 \mathrm{~volt}$

Topic: Physics, Electromagnetic Induction, JEE Main
The magnetic flux linked with a vector area $\vec{A}$ in a uniform magnetic field $\vec{B}$ is

Options:
A $\vec{B} \times \vec{A}$
B $A B$
C $\vec{B} \cdot \vec{A}$
D ${ }^{\frac{B}{A}}$

The correct answer is:
$\vec{B} \cdot \vec{A}$

Solution:
Magnetic flux is a measurement of the total magnetic field which passes through a given area.
If we chose a simple flat surface with surface area $\mathrm{A}$ as our test area and there is an angle $\theta$ between the normal to the surface and a magnetic field $B$, then the magnetic flux is,
$\Phi=B A \cos \theta=\operatorname{dot}$ product between the two vectors
hence, option 3 is correct

Topic: Physics, Electromagnetic Induction, JEE Main
The magnetic flux through a circuit of resistance $R$ changes by an amount $\Delta \phi$ in a time $\Delta t$. Then the total quantity of electric charge $Q$ that passes any point in the circuit during the time $\Delta \mathrm{t}$ is represented by:

Options:
A $\mathrm{Q}=\frac{1}{R} \cdot \frac{\Delta \phi}{\Delta t}$
B $Q=\frac{\Delta \phi}{R}$
C $Q=\frac{\Delta \phi}{\Delta t}$
D $Q=R \cdot \frac{\Delta \phi}{\Delta t}$

The correct answer is:
$Q=\frac{\Delta \phi}{R}$

Solution:
According to the question on induced emf is:
$$
\begin{aligned}
& \text { and current } I=\frac{Q}{\Delta t}=\frac{\Delta \phi}{\Delta t} \cdot \frac{1}{R} \\
& \Rightarrow \quad q=\frac{\Delta \phi}{R}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, JEE Main
The magnetic flux through a circuit of resistance R changes by an amount Î Ï in a time Î t . Then the total quantity of electric charge Q , that passes through any point in the circuit during this Î t time is, represented by

Options:
A Q = 1 R Δ Ï Δ t
B Q = Î Ï R
C Q = Δ Ï Δ t
D Q = R Δ Ï Δ t

The correct answer is:
Q = Δ ϕ R

Solution:
Induced emf is given by Îµ = Î Ï Î t
Current, I = Q Î t = Î Ï Î t Ã 1 R [where Q is total charge passed in time Î t ]
â Q = Î Ï R

Topic: Physics, Electromagnetic Induction, JEE Main
The magnetic flux through a coil perpendicular to its plane is varying according to the relation ϕ=5t3+4t2+2t-5 Weber. If the resistance of the coil is 5 ohm, then the induced current through the coil at t=2 s will be,

Options:
A 15.6 A
B 16.6 A
C 17.6 A
D 18.6 A

The correct answer is:
15.6 A

Solution:
Given, ϕ=5t3+4t2+2t-5

Magnitude of induced emf ε=dϕdt=15t2+8t+2

At, t=2 s, ε=60+16+2=78 V

Current, i=εR=785=15.6 A

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2022 (26 Jun Shift 1)
The magnetic force acting on a charged particle of charge $-2 \mu \mathrm{c}$ in a magnetic field of $2 \mathrm{~T}$ acting in $\mathrm{y}$ direction, when the particle velocity is $(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \times 10^6 \mathrm{~ms}^{-1}$, is :

Options:
A $8 \mathrm{~N}$ in $\mathrm{z}$ - direction
B $8 \mathrm{~N}$ in $\mathrm{z}-$ direction
C $4 \mathrm{~N}$ in $\mathrm{z}$ - direction
D $8 \mathrm{~N}$ in $\mathrm{y}-$ direction

The correct answer is:
$8 \mathrm{~N}$ in $\mathrm{z}-$ direction

Solution:
$$
\begin{aligned}
\overrightarrow{\mathrm{F}} & =\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \\
& =-2 \times 10^{-6}\left[(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \times 10^6 \times 2 \hat{\mathrm{j}}\right] \\
& =-(8 \mathrm{~N}) \hat{\mathrm{k}}
\end{aligned}
$$

Topic: Physics, Magnetic Properties of Matter, JEE Main
The magnetic force acting on a charged particle of charge $-2 \mu \mathrm{c}$ in a magnetic field of $2 \mathrm{~T}$ acting in $\mathrm{y}$ direction, when the particle velocity is $(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \times 10^6 \mathrm{~ms}^{-1}$, is :

Options:
A $8 \mathrm{~N}$ in $\mathrm{z}$ - direction
B $8 \mathrm{~N}$ in $\mathrm{z}-$ direction
C $4 \mathrm{~N}$ in $\mathrm{z}$ - direction
D $8 \mathrm{~N}$ in $\mathrm{y}-$ direction

The correct answer is:
$8 \mathrm{~N}$ in $\mathrm{z}-$ direction

Solution:
$$
\begin{aligned}
\overrightarrow{\mathrm{F}} & =\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \\
& =-2 \times 10^{-6}\left[(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \times 10^6 \times 2 \hat{\mathrm{j}}\right] \\
& =-(8 \mathrm{~N}) \hat{\mathrm{k}}
\end{aligned}
$$

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2009 (Mains)
The magnetic force acting on charged particle of charge $2 \mu \mathrm{C}$ in magnetic field of $2 T$ acting in $y-$ direction, when the particle velocity is $(2 \hat{i}+3 \hat{j}) \times 10^6 \mathrm{~ms}^{-1}$ is

Options:
A
$8 \mathrm{~N}$ in z-direction

B
$8 \mathrm{~N}$ in y-direction

C
$4 \mathrm{~N}$ in y-direction

D
$4 \mathrm{~N}$ in z-direction

The correct answer is:
$8 \mathrm{~N}$ in z-direction

Solution:
$$
\begin{aligned}
& \text { } \vec{F}=q(\vec{V} \times \vec{B}) \\
& =2 \times 10^{-6}\left[(2 \hat{i}+3 \hat{j}) \times 10^6 \times 2 \hat{j}\right] \\
&
\end{aligned}
$$
$=2 \times 4 \hat{k}=8 N$ in Z-direction.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2012 (12 May Online)
The magnetic induction due to an infinitely long straight wire carrying a current $i$ at a distance $r$ from wire is given by

Options:
A $|\boldsymbol{B}|=\left(\frac{\mu_0}{4 \pi}\right) \frac{2 i}{r}$
B $|\boldsymbol{B}|=\left(\frac{\mu_0}{4 \pi}\right) \frac{r}{2 i}$
C $|\boldsymbol{B}|=\left(\frac{4 \pi}{\mu_0}\right) \frac{2 i}{r}$
D $|\boldsymbol{B}|=\left(\frac{4 \pi}{\mu_0}\right) \frac{r}{2 i}$

The correct answer is:
$|\boldsymbol{B}|=\left(\frac{\mu_0}{4 \pi}\right) \frac{2 i}{r}$

Solution:
Correct option is A. $|B|=\left(\frac{\mu_0}{4 \pi}\right) \frac{2 i}{r}$
The magnetic field due to a long current carrying conductor is given by:
$|B|=\left(\frac{\mu_0}{4 \pi}\right) \frac{2 i}{r}$

Topic: Physics, Magnetic Effects of Current, JEE Main
The magnetic induction in air at a point $1 \mathrm{~cm}$ away from a long wire that carries a current of $1 A$, will b

Options:
A $1 \times 10^{-5} \mathrm{~T}$
B $2 \times 10^{-5} \mathrm{~T}$
C $3 \times 10^{-5} \mathrm{~T}$
D $4 \times 10^{-5} \mathrm{~T}$

The correct answer is:
$2 \times 10^{-5} \mathrm{~T}$

Solution:
$B=\frac{\mu_0}{4 \pi} \times \frac{2 i}{r}=10^{-7} \times \frac{2 \times 1}{10^{-2}}=2 \times 10^{-5}$ Tesla

Topic: Physics, Magnetic Effects of Current, JEE Main
The magnetic induction in the region between the pole faces of an electromagnet is $0.7 \mathrm{weber} / \mathrm{m}^2$. The induced e.m.f. in a straight conductor $10 \mathrm{~cm}$ long, perpendicular to $B$ and moving perpendicular both to magnetic induction and its own length with a velocity $2 \mathrm{~m} / \mathrm{sec}$ is

Options:
A $0.08 \mathrm{~V}$
B $0.14 \mathrm{~V}$
C $0.35 \mathrm{~V}$
D $0.07 \mathrm{~V}$

The correct answer is:
$0.14 \mathrm{~V}$

Solution:
$e=B v l \Rightarrow e=0.7 \times 2 \times\left(10 \times 10^{-2}\right)=0.14 \mathrm{~V}$

Topic: Physics, Electromagnetic Induction, JEE Main
The magnetic intensity at the centre of a long current carrying solenoid is found to be 1.6 ×103 A m-1. If the number of turns is 8 per cm, then the current flowing through the solenoid is ______A.

Options:

The correct answer is:
2

Solution:
The magnetic intensity H is defined as the product of the number of turns per unit length in a coil n and the current that it carries I.

The magnetic field of a solenoid is given by B=μ0ni

H=Bμ0=μ0niμ0⇒H=ni ...(i)

The given data is

H=1.6×103 A m-1

n=8 per cm

Substituting the values in equation (i)

⇒1.6×1000=81100×i

⇒i=2 A

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The magnetic lines of force inside a bar magnet

Options:
A
are from north-pole to south-pole of the magnet

B
do not exist

C
depend upon the area of cross-section of the bar magnet

D
are from south-pole to north-pole of the magnet

The correct answer is:
are from south-pole to north-pole of the magnet

Solution:
No Solution Available

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2003
The magnetic moment is measured in Bohr Magneton (BM). Spin only magnetic moment of Fe in FeH2O63+ and FeCN63- complexes respectively is:

Options:
A 6.92 B.M. in both
B 3.87 B.M. and1.732 B.M.
C 5.92 B.M. and 1.732 B.M.
D 4.89 B.M. and 6.92 B.M.

The correct answer is:
5.92 B.M. and 1.732 B.M.

Solution:
The spin only magnetic moment can be calculated as follows

μ=nn+2BM

n=Number of unpaired electrons

FeH2O63+=d5(High spin complex)=5.unpaired electrons
μ=5×7=35
=5.92 B.M.
FeCN63+=d5(Low spin complex)=1
μ=1×3=3
=1.732 B.M.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (11 Apr Shift 2)
The magnetic moment $\left(\mathrm{m}_{\mathrm{orb}}\right)$ of a revolving electron around the nucleus varies with the principal quantum number (n) as

Options:
A $\mathrm{m}_{\text {orb }} \propto \mathrm{n}^2$
B $\mathrm{m}_{\mathrm{orb}} \propto \frac{1}{\mathrm{n}^2}$
C $\mathrm{m}_{\mathrm{orb}} \propto \frac{1}{\mathrm{n}}$
D $\mathrm{m}_{\text {orb }} \propto \mathrm{n}$

The correct answer is:
$\mathrm{m}_{\text {orb }} \propto \mathrm{n}$

Solution:
Orbital magnetic moment can be defined as,
$\mathrm{m}_{\text {orb }}=\mathrm{iA}$
where, $\mathrm{i}=\frac{\mathrm{e}}{\mathrm{T}}, \mathrm{r}$ is the radius of the Bohr orbit, $\mathrm{A}=\pi \mathrm{r}^2$ is the area and $\mathrm{T}$ is the time period of uniform circular motion.
Now, we can make use of the Bohr's hypothesis about angular momentum:
On plugging in above into equation (1),
$\mathrm{m}_{\mathrm{orb}}=\frac{\mathrm{neh}}{4 \pi \mathrm{m}}$
Orbital magnetic moments of an electron in Bohr orbit is given by,
$\begin{aligned} & \mathrm{m}_{\text {orb }}=\mathrm{n}\left(\frac{\mathrm{eh}}{4 \pi \mathrm{m}}\right) \\ & \therefore \mathrm{m}_{\text {orb }} \propto \mathrm{n}\end{aligned}$

Topic: Physics, Atomic Physics, JEE Main
The magnetic moment of a bar magnet is $0.5 \mathrm{Am}^2$. It is suspended in a uniform magnetic field of $8 \times 10^{-2} \mathrm{~T}$. The work done in rotating it from its most stable to most unstable position is:

Options:
A $8 \times 10^{-2} \mathrm{~J}$
B $4 \times 10^{-2} \mathrm{~J}$
C Zero
D $16 \times 10^{-2} \mathrm{~J}$

The correct answer is:
$8 \times 10^{-2} \mathrm{~J}$

Solution:
At stable equilibrium
$\mathrm{U}=-\mathrm{mB} \cos 0^{\circ}=-\mathrm{mB}$
At unstable equilibrium
$\begin{aligned}
& \mathrm{U}=-\mathrm{mB} \cos 180^{\circ}=+\mathrm{mB} \\
& \mathrm{W}=\Delta \mathrm{U} \\
& \text { W.D. }=2 \mathrm{mB} \\
& =2(0.5) 8 \times 10^{-2}=8 \times 10^{-2} \mathrm{~J}
\end{aligned}$

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2024 (04 Apr Shift 2)
The magnetic moment of a diamagnetic atom is

Options:
A much greater than one
B one
C between zero and one
D equal to zero

The correct answer is:
equal to zero

Solution:
Atoms which have paired electrons have the magnetic moment zero. Because diamagnetism is the intrinsic property of every material and it is generated due to mutual interaction between the applied magnetic field and orbital motion of electrons.
So, the magnetic moment of a diamagnetic atom is equal to zero.

Topic: Physics, Magnetic Properties of Matter, JEE Main
The magnetic moment of a diamagnetic atom is

Options:
A much greater than one
B one
C between zero and one
D equal to zero

The correct answer is:
equal to zero

Solution:
Atoms which have paired electrons have the magnetic moment zero. Because diamagnetism is the intrinsic property of every material and it is generated due to mutual interaction between the applied magnetic field and orbital motion of electrons.
So, the magnetic moment of a diamagnetic atom is equal to zero.

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2010 (Mains)
The magnetic moment of a magnet of length $10 \mathrm{~cm}$ and pole strength $40 \mathrm{Am}$ will be

Options:
A $1.6 \mathrm{Am}^2$
B $0.4 \mathrm{Am}^2$
C $20 \mathrm{Am}^2$
D $8.0 \mathrm{Am}^2$

The correct answer is:
$0.4 \mathrm{Am}^2$

Solution:
$\mathrm{M}=\mathrm{mL}=4 \times 10 \times 10^{-2}=0.4 \mathrm{~A} \times \mathrm{m}^2$

Topic: Physics, Magnetic Properties of Matter, JEE Main
The magnetic moment of a transition metal compound has been calculated to be 3.87 BM. The metal ion is

Options:
A Cr2+
B Mn2+
C V2+
D Ti2+

The correct answer is:
V2+

Solution:
The magnetic moment, μs​=n(n+2)​BM where n is the number of unpaired electrons.

Cr+2 : Ar, 3d4, 4s° n=4, μ=4(4+2)=24=4.89 BMMn+2 : Ar, 3d5, 4s° n=5, μ=5(5+2)=35=5.91 BMV+2 : Ar, 3d3, 4s° n=3, μ=3(3+2)=15=3.87 BMTi+2 : Ar, 3d2, 4s° n=2, μ=2(2+2)=8=2.82 BM

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2023 (24 Jan Shift 1)
The magnetic moment of a transition metal ion is $\sqrt{15}$ BM. Therefore, the number of unpaired electrons present in it, is

Options:
A 3
B 4
C 1
D 2

The correct answer is:
3

Solution:
$\begin{array}{lrl}\mu=\sqrt{n(n+2)} & & \\ \Rightarrow & \sqrt{15} & =\sqrt{n(n+2)} \\ \therefore & n & =3\end{array}$

Topic: Chemistry, Structure of Atom, JEE Main
The magnetic moment of an electron e revolving in an orbit around nucleus with an orbital angular momentum is given by

Options:
A μ→L=eL2 m
B μ→L=-eL2 m
C μ→L=-eL→m
D μ→L=2eL→m

The correct answer is:
μ→L=-eL2 m

Solution:
When a charged body is rotating in a circle, the ratio of magnetic moment and angular momentum is given by,

μ→L→=q2m

The charge of electron q=-e

μ→=-e2mL→

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (25 Jul Shift 1)
The magnetic moment of an octahedral homoleptic Mn(II) complex is 5.9 B.M. . The suitable ligand for this complex is:

Options:
A CN-
B CO
C Ethylenediamine
D NCS-

The correct answer is:
NCS-

Solution:
Mn2+ will have configuration =3d54s0
μ=5.93=nn+2
or, n=5 as 35=5.93
If has 5 unpaired electron which are not getting paired up. This implies that the ligand paired up, and the ligand is a weak field ligand. Among the given options, weak ligand is NCS-

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The magnetic moment of the complex anion $\left[\mathrm{Cr}(\mathrm{NO})\left(\mathrm{NH}_3\right)(\mathrm{CN})_4\right]^{2-}$ is :

Options:
A
$5.91 \mathrm{BM}$

B
$3.87 \mathrm{BM}$

C
$1.73 \mathrm{BM}$

D
$2.82 \mathrm{BM}$

The correct answer is:
$2.82 \mathrm{BM}$

Solution:
$\operatorname{In}\left[\mathrm{Cr}(\mathrm{NO})\left(\mathrm{NH}_3\right)(\mathrm{CN})_4\right]^{2-}$, $\mathrm{Cr}^{2+}\left(\mathrm{d}^4\right)$ is given as :
i.e., 2 unpaired electrons
$$
\mu=\sqrt{2(2+2)}=\sqrt{8}=2.82
$$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2013 (23 Apr Online)
The magnetic moment (spin only) of $\left[\mathrm{NiCl}_4\right]^{2-}$ is

Options:
A
$5.46 \mathrm{BM}$

B
$2.83 \mathrm{BM}$

C
$1.41 \mathrm{BM}$

D
$1.82 \mathrm{BM}$

The correct answer is:
$2.83 \mathrm{BM}$

Solution:
In $\left[\mathrm{NiCl}_4\right]^{2-}, \mathrm{n}=2$
$\begin{aligned} \mu & =\sqrt{n(n+2)} \mathrm{BM} \\ & =\sqrt{2(2+2)}=2.82 \mathrm{BM}\end{aligned}$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2011
The magnetic moments associated with two closely wound circular coils A and B of radius rA=10 cm and rB=20 cm respectively are equal if: (Where NA, IA and NB, IB are number of turn and current of A and B respectively)

Options:
A 2NAIA=NBIB
B NA=2NB
C NAIA=4NBIB
D 4NAIA=NBIB

The correct answer is:
NAIA=4NBIB

Solution:
As magnetic moment of both the coil is the same,

μ1=μ2

⇒NAIAπrA2=NBIBπrB22⇒NAIA0.102=NBIB0.202⇒IANA=4IBNB

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (30 Jan Shift 1)
The magnetic needle of a vibration magnetometer makes 12 oscillations per minute in the horizontal component of earth's magnetic field. When an external short bar magnet is placed at some distance along the axis of the needle in the same line, it makes 15 oscillations per minute. If the poles of the bar magnet are interchanged, the number of oscillations it makes per minute is n. What is the value of n?

Options:

The correct answer is:
63

Solution:
Ist case :
n=12πMBHI
⇒ n∝BH ⇒ n1n2=BHBH+BH1
⇒ 1215=BHBH+BH1
⇒ BH1=916BH
IInd case :
n2n3=BH+BH1BH-BH1
⇒ 15n3=BH+916BHBH-916BH
∴ n3=63⇒n=63

Topic: Physics, Magnetic Properties of Matter, JEE Main
The magnetic needle of a vibration magnetometer makes 12 oscillations per minute in the horizontal component of earth's magnetic field. When an external short bar magnet is placed at some distance along the axis of the needle in the same line, it makes 15 oscillations per minute. If the poles of the bar magnet are interchanged, the number of oscillations it makes per minute is

Options:
A $\sqrt{61}$
B $\sqrt{63}$
C $\sqrt{65}$
D $\sqrt{67}$

The correct answer is:
$\sqrt{63}$

Solution:
$\begin{array}{ll}
\text {Ist case } & n=\frac{1}{2 \pi} \sqrt{\frac{M H}{I}} \\
\Rightarrow & n \propto \sqrt{H} \Rightarrow \frac{n_1}{n_2}=\sqrt{\frac{H}{H+H_1}} \\
\Rightarrow \quad & \frac{12}{15}=\sqrt{\frac{H}{H+H_1}} \Rightarrow H_1=\frac{9}{16} H
\end{array}$
IInd case
$\begin{aligned}
\frac{n_2}{n_3} & =\sqrt{\frac{H+H_1}{H-H_1}} \Rightarrow \frac{15}{n_3}=\sqrt{\frac{H+\frac{9}{16} H}{B-\frac{9}{16} H}} \\
\because \quad n_3 & =\sqrt{63}
\end{aligned}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
The magnetic needle of a vibration magnetometer makes 12 oscillations per minute in the horizontal component of earth's magnetic field. When an external short bar magnet is placed at some distance along the axis of the needle in the same line, it makes 15 oscillations per minute. If the poles of the bar magnet are interchanged, the number of oscillations it makes per minute is

Options:
A $\sqrt{61}$
B $\sqrt{63}$
C $\sqrt{65}$
D $\sqrt{67}$

The correct answer is:
$\sqrt{63}$

Solution:
$\begin{array}{ll}
\text { Ist case } & n=\frac{1}{2 \pi} \sqrt{\frac{M H}{I}} \\
\Rightarrow & n \propto \sqrt{H} \Rightarrow \frac{n_1}{n_2}=\sqrt{\frac{H}{H+H_1}} \\
\Rightarrow \quad & \frac{12}{15}=\sqrt{\frac{H}{H+H_1}} \Rightarrow H_1=\frac{9}{16} H
\end{array}$
IInd case
$\begin{aligned}
\frac{n_2}{n_3} & =\sqrt{\frac{H+H_1}{H-H_1}} \Rightarrow \frac{15}{n_3}=\sqrt{\frac{H+\frac{9}{16} H}{B-\frac{9}{16} H}} \\
\because \quad n_3 & =\sqrt{63}
\end{aligned}$

Topic: Physics, Magnetic Properties of Matter, JEE Main