datavorous/entrance-exam-dataset huggingface dataset:
The lower 24 mportion of a $50 \mathrm{~m}$ tall tower is painted green and the remaining portion red. What is the distance of a point on the ground from the base of the tower where the two different portions of the tower subtend equal angles?

Options:
A $60 \mathrm{~m}$
B $72 \mathrm{~m}$
C $90 \mathrm{~m}$
D $\begin{array}{ll}\text { } 120 \mathrm{~m} & {}\end{array}$

The correct answer is:
$\begin{array}{ll}\text { } 120 \mathrm{~m} & {}\end{array}$

Solution:
Let the distance, be $\mathrm{x}$, and angle $\mathrm{APB}=\theta$, then $\angle \mathrm{BPC}=\angle \mathrm{APC}=\theta / 2$
In triangle $\Delta \mathrm{APB}$,
$\tan \theta=\frac{\mathrm{AB}}{\mathrm{x}}=\frac{50}{\mathrm{x}}$
.(1)
and in triangle $\mathrm{APC}$
$\tan \frac{\theta}{2}=\frac{\mathrm{AC}}{\mathrm{x}}=\frac{24}{\mathrm{x}}$
$\tan \theta=\frac{2 \tan \frac{\theta}{2}}{1-\tan ^{2} \frac{\theta}{2}}$
Putting the value of $\tan \theta$ and $\tan \frac{\theta}{2}$
From equation (1) and (2) in equation (3),
$\frac{50}{x}=\frac{2 \times \frac{24}{x}}{1-\left(\frac{24}{x}\right)^{2}}$
or, $\frac{50}{x}=\frac{48}{x} \times \frac{x^{2}}{x^{2}-(24)^{2}}$
or, $50\left\{\mathrm{x}^{2}-(24)^{2}\right\}=48 \mathrm{x}^{2}$
or, $50 \mathrm{x}^{2}-50 \times(24)^{2}=48 \mathrm{x}^{2}$
or, $2 \mathrm{x}^{2}=(24)^{2} \times 50$
$\mathrm{x}=25 \times(24)^{2}$
$\mathrm{x}=|5 \times 24|=120 \mathrm{~m}$
Such a point can exist on the either side of the tower.

Topic: Mathematics, Heights and Distances, NDA, NDA 2007 (Phase 1)
The lower end of a glass capillary tube is dipped in water. Water rises to a height of 9 cm. The tube is then broken at a height of 5 cm. The height of the water column and angle of contact will be

Options:
A 5 cm, cos-1⁡59
B 4 cm, cos-1⁡54
C 5 cm, cos-1⁡95
D 5 cm, cos-1⁡67

The correct answer is:
5 cm, cos-1⁡59

Solution:
When a capillary tube is broken at a height 5 cm, meaning that water will rise to height 5 cm, hence h2=5 cm.
Since, we know, for capillarity phenomenon h=2Scosθrρg, where symbols have their usual meaning.
or hcosθ= constant
Thus, h1cosθ1=h2cosθ2
⇒9cosθ=5cosθ2 [for glass θ1=0 ]
⇒cosθ2=59 ∵ cos0o=1
∴θ2=cos-159

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The lower stability of ethyl anion compared to methyl anion and the higher stability of ethyl radical compared to methyl radical, respectively, are due to

Options:
A +I -effect of the methyl group in ethyl anion, σ→p -orbital conjugation in ethyl radical
B -I -effect of the methyl group in ethyl anion and σ→σ* conjugation in ethyl radical
C +I effect of the methyl group in both cases
D +I - effect of the methyl group in ethyl anion and σ→σ* conjugation in ethyl radical

The correct answer is:
+I -effect of the methyl group in ethyl anion, σ→p -orbital conjugation in ethyl radical

Solution:
The lower stability of ethyl anion CH3CH2- Compared to methyl anion CH3- is because of +I -effect of methyl group of ethyl anion. The higher stability of ethyl radical compared to methyl radical is due to σ-p -orbital conjugation which is known as hyper conjugation in ethyl radical.

Topic: Chemistry, General Organic Chemistry, JEE Main
The lowest integer which is greater than 1+11010010100 is

Options:
A 3
B 4
C 2
D 1

The correct answer is:
3

Solution:
Consider P=1+11010010100,

Let x=10100

⇒P=1+1xx

⇒P=1+x1x+xx-1⌊2·1x2+xx-1x-233·1x3+…

(upto 10100+1 terms)

⇒P=1+1+1⌊2-12x+1⌊3-…+… so on

⇒P=2+(Positive value less than 1⌊2+1⌊3+1⌊4+… )

Also e=1+1⌊1+1⌊2+1⌊3+1⌊4+…

⇒1⌊2+1⌊3+1⌊4+…=e-2

⇒P=2+ (Positive value less than e-2)

⇒P∈2,3

⇒ Least integer value of P is 3.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (25 Jul Shift 2)
The lowest resistance which can be obtained by connecting 10 resistors each of $1 / 10 \mathrm{ohm}$ is

Options:
A $1 / 250 \Omega$
B $1 / 200 \Omega$
C $1 / 100 \Omega$
D $1 / 10 \Omega$

The correct answer is:
$1 / 100 \Omega$

Solution:
Lowest resistance will be in the case when all the resistors are connected in parallel.
$\frac{1}{R}=\frac{1}{0.1}+\frac{1}{0.1} \ldots . .10$ times
$\frac{1}{R}=10+10 \ldots . . .10$ times
$\frac{1}{R}=100$ i.e. $R=\frac{1}{100} \Omega$

Topic: Physics, Current Electricity, JEE Main
The lysosome contains hydrolytic enzymes. For the functioning of enzymes, the pH inside the lysosome is

Options:
A independent of the H+ ion concentration.
B decreased by active pumping of H+ to the cytoplasm.
C increased by simple diffusion of H+ from the cytoplasm.
D dependent on the K+ ion concentration of the cytoplasm.

The correct answer is:
decreased by active pumping of H+ to the cytoplasm.

Solution:
Lysosomes are formed by budding off vesicles from the Golgi apparatus. They are bounded by a single membrane about 500 nm and are involved in intracellular digestion, containing the enzymes needed to digest all types of macromolecules. The enzymes present in such vesicles belong to acid hydrolases: The acidic condition is maintained by pumping protons into the interior of the lysosomes from the cytoplasm. (Lysosomes contain hydrolytic enzymes that require acidic pH to function, for the functioning of these enzymes pH is decreased by actively pumping H+ to the cytoplasm).

Topic: Biology, Cell - The Unit of Life, NEET
The machine as shown has 2 rods of length 1 m connected by a pivot at the top. The end of one rod is connected to the floor by a stationary pivot and the end of the other rod has roller that rolls along the floor in a slot. As the roller goes back and forth, a 2 kg weight moves up and down. If the roller is moving towards right at a constant speed, the weight moves up with a :

Options:
A Speed which is 34th of that of the roller when the weight is 0.4 m above the ground
B Constant speed
C Decreasing speed
D Increasing speed

The correct answer is:
Decreasing speed

Solution:
vA=dxdt=d2lcosθdt=-2lsinθdθdt and

Vertical velocity of B, vB=dydt=dlsinθdt=lcosθdθdt

vAvB=-2tanθ⇒vB=-vAcotθ2

θ→increase

cotθ→decrease

Hence, vB→decrease.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2017 (09 Apr Online)
The magnetic behaviour of Li2O, Na2O2 and KO2, respectively, are

Options:
A diamagnetic, paramagnetic and diamagnetic
B paramagnetic, paramagnetic and diamagnetic
C paramagnetic, diamagnetic and paramagnetic
D diamagnetic, diamagnetic and paramagnetic

The correct answer is:
diamagnetic, diamagnetic and paramagnetic

Solution:
In the structure Li2O each lithium atom loses one electron to form 2 cations Li+ and oxygen atom gains two electrons to form the anion O2-. It has no unpaired electrons. So, it is diamagnetic in nature.

The peroxide O22- present in Na2O2 has 4 electron in π*2p, thus all are paired. So, it is diamagnetic in nature.

In the structure KO2 oxygen is present in O2-. The number of electrons present is 17, where the where the sixteen electrons are from two oxygen atoms and one from the negative charge. There is one unpaired electron present in the last orbital. Hence, it is paramagnetic in nature.

Li2O→ diamagnetic

Na2O2→ diamagnetic

KO2→ paramagnetic

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (29 Jan Shift 1)
The magnetic dipole moment of a short magnetic dipole at a distant point along the equator of magnet has a magnitude of ‘X’ in SI units. If the distance between point and the magnet is halved then the magnitude of dipole moment will be

Options:
A 2X
B 12X
C X
D 18X

The correct answer is:
X

Solution:
The magnetic dipole moment is the product of either of pole strength and the magnetic length of dipole. Thus, it is independent of the distance of point at which it is measured. So, it remains unchanged, if the distance between point and the magnet halved.

Topic: Physics, Magnetic Effects of Current, MHT CET, MHT CET 2019 (Shift 2)
The magnetic energy stored in an inductor of inductance 4 μH carrying a current of 2 A is:

Options:
A 8 μJ
B 4 μJ
C 4 mJ
D 8 mJ

The correct answer is:
8 μJ

Solution:
Magnetic energy stored in an inductor is given by,

U=12Li2=12×4×10-6×22=8×10-6 J=8 μJ

Topic: Physics, Electromagnetic Induction, JEE Main
The magnetic field associated with a light wave is given, at the origin, by B=B0 sinâ¡3.14Ã107ct+sinâ¡6.28Ã107ct. If this light falls on a silver plate having a work function of 4.7eV, what will be the maximum kinetic energy of the photoelectrons?
(c=3Ã108âmâsâ1,h=6.6Ã10â34âJâs)

Options:
A 6.82 eV
B 7.72 eV
C 12.5 eV
D 8.52 eV

The correct answer is:
7.72 eV

Solution:
B=B0sin⁡3.14×107ct+sin⁡6.28×107ct
Maximum KE will associate with the maximum frequency
fmax⁡ =6.28×107c2π=107×3×108 Hz =3×1015 Hz
KEmax=hfmax-ϕ
=6.63×10-34×3×10151.6×10-19-4.7 eV
=7.72 eV

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The magnetic field associated with a light wave is given, at the origin, by B=B0 sinâ¡3.14Ã107ct+sinâ¡6.28Ã107ct. If this light falls on a silver plate having a work function of 4.7eV, what will be the maximum kinetic energy of the photoelectrons?
(c=3Ã108âmâsâ1,h=6.6Ã10â34âJâs)

Options:
A 6.82 eV
B 7.72 eV
C 12.5 eV
D 8.52 eV

The correct answer is:
7.72 eV

Solution:
B=B0sin⁡3.14×107ct+sin⁡6.28×107ct
Maximum KE will associate with the maximum frequency
fmax⁡ =6.28×107c2π=107×3×108 Hz =3×1015 Hz
KEmax=hfmax-ϕ
=6.63×10-34×3×10151.6×10-19-4.7 eV
=7.72 eV

Topic: Physics, Dual Nature of Matter, JEE Main
The magnetic field at P in the arrangement shown is

Options:
A μ0i2πd1-12⊗
B 2μ0i2πd⊗
C μ0i2πd⊗
D μ0i22πd1+12⊗

The correct answer is:
μ0i2πd1-12⊗

Solution:
xd=sin45° ⇒x=d2
we can extend the wire and then subtract the field of the extended part.
B1=μ04πId/2sin0°+sin90°-μ04πId/2sin0°+sin45°
=μ04π2Id1-12=μ0I22 πd1-12⊗
B1⊗,B2⊗⇒B=B1+B2⊕=2B1⊗=μ0I2 πd1-12⊗

Topic: Physics, Magnetic Effects of Current, JEE Main
The magnetic field at the center of current carrying circular loop is B1. The magnetic field at a distance of 3 times radius of the given circular loop from the center on its axis is B2. The value of B1B2 will be

Options:
A 9:4
B 12:5
C 8:1
D 5:3

The correct answer is:
8:1

Solution:
Magnetic field at the centre of the loop i.e. B1=μ0i2R

Magnetic field on the axis of the loop is,

B2=μ0iR22R2+x232=μ0i×R22R2+3R232=μ0i×R22×8R3=μ0i2×8R

Therefore, the ratio B1B2=81

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (28 Jul Shift 2)
The magnetic field at the centre of a circular coil of radius $r$ carrying current $/$ is $B_1$. The field at the centre of another coil of radius $2 r$ carrying same current $/$ is $B_2$. The ratio $\frac{B_1}{B_2}$ is

Options:
A $\frac{1}{2}$
B 1
C 2
D 4

The correct answer is:
2

Solution:
$B \propto \frac{1}{r} \Rightarrow \frac{B_1}{B_2}=\frac{r_2}{r_1}=\frac{2 r}{r}=2$

Topic: Physics, Magnetic Effects of Current, JEE Main
The magnetic field at the centre of a circular coil of radius r, due to current I flowing through it, is B. The magnetic field at a point along the axis at a distance r2 from the centre is :

Options:
A B2
B 2B
C 253 B
D 233 B

The correct answer is:
253 B

Solution:
Magnetic field at the centre of the circle is, B=μ0i2R

Magnetic field at the point on the axis is,

B2=μ0iR22R2+x232⇒B2=μ0iR22R2+R2232⇒B2=μ0i25432R=253B

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (24 Jun Shift 1)
The magnetic field at the centre of a current-carrying loop of radius 0.1 m is 55 times that at a point along its axis. What is the distance (in m) of this point from the centre of the loop?

Options:

The correct answer is:
0.2

Solution:
We know that,
BcentreBaxis=1+x2r23/2
Given that, Bcentre=55 Baxis
BcentreBaxis=55
â´ 55=1+x20.123/2
On squaring both sides, we get
25Ã5=1+x20.123
1253=1+x20.12
â 0.01+x2=0.05
â x2=0.05-0.01
â x2=0.04
â x=0.2 m

Topic: Physics, Magnetic Effects of Current, JEE Main
The magnetic field at the centre of a wire loop formed by two semicircular wires of radii R1=2π m and R2=4π m carrying current I=4 A as per figure given below is α×10-7 T. The value of α is _____. (Centre O is common for all segments)

Options:

The correct answer is:
3

Solution:
From the figure, it is evident that the magnetic field at O due to the straight portion of the loop is zero, only the contributions from the circular parts survive.

Also, the direction of the magnetic field due to the circular loops can be determined by the right-hand corkscrew rule.

Hence, the net magnetic field at O can be calculated as follows:

B=μ0i2R2π2π⊗+μ0i2R1π2π⊗=μ0i4R2+μ0i4R1⊗=4π×10-7×44×4π+4π×10-7×44×2π⊗ T=3×10-7⊗ T

Hence, α=3.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The magnetic field at the point of intersection of diagonals of a square loop of side L carrying a current I is

Options:
A μ0IπL
B 2μ0IπL
C 2μ0IπL
D 22μ0IπL

The correct answer is:
22μ0IπL

Solution:
From Biot-Savart's law, magnetic field due to current carrying conductor is

B= μ04πir sin⁡ϕ1+sin⁡ϕ2
Where, ϕ1=45°,ϕ2=45°

B=μ04π .iL212+12=μ0i22πL
Since there are four sides, the total magnetic field is

Bnet=4B=22μ0iπL

Topic: Physics, Magnetic Effects of Current, JEE Main
The magnetic field at the point of intersection of diagonals of a square wire loop of side \(\mathrm{L}\) carrying a current \(\mathrm{I}\) is

Options:
A \(\frac{\mu_0 \mathrm{I}}{\pi \mathrm{L}}\)
B \(\frac{2 \mu_0 \mathrm{I}}{\pi \mathrm{L}}\)
C \(\frac{\sqrt{2} \mu_0 I}{\pi \mathrm{L}}\)
D \(\frac{2 \sqrt{2} \mu_{\mathrm{o}} \mathrm{I}}{\pi \mathrm{L}}\)

The correct answer is:
\(\frac{2 \sqrt{2} \mu_{\mathrm{o}} \mathrm{I}}{\pi \mathrm{L}}\)

Solution:
\(\begin{aligned} & \text {Hints: } B=4\left[\frac{\mu_0}{4 \pi} \frac{I}{\left(\frac{\mathrm{L}}{2}\right)}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)\right] \\ & =\frac{\mu_{\mathrm{o}}}{\pi} \frac{2 \mathrm{I}}{\mathrm{L}} \cdot \frac{2}{\sqrt{2}} ; \quad B=\frac{\mu_{\mathrm{o}}}{\pi} \frac{2 \sqrt{2} \mathrm{I}}{\mathrm{L}}\end{aligned}\)

Topic: Physics, Magnetic Effects of Current, JEE Main
The magnetic field B crossing normally a square metallic plate of area 4 m2 is changing with time as shown in figure. The magnitude of induced emf in the plate during t=2 s to t=4 s, is _____ mV.

Options:

The correct answer is:
8

Solution:
From the graph, it can be seen that the value of the magnetic field at t=2 s is 4 mT, whereas the magnetic field at t=4 s is 8 mT.

The formula to calculate the magnitude of the induced emf is given by

εind =dϕdt=ddtBA=AdBdt ...1

dBdt is the slope of the given graph.

Hence, the value of the induced emf can be calculated as follows:

εind =4 m2×8 mT-4 mT4 s-2 s=8 mV

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The magnetic field $d \vec{B}$ due to a small current element $d \vec{i}$ at a distance $\vec{r}$ and element carrying current $i$ is, or
Vector form of Biot-savart's law is

Options:
A $d \vec{B}=\frac{\mu_0}{4 \pi} i\left(\frac{d \vec{l} \times \vec{r}}{r}\right)$
B $d \vec{B}=\frac{\mu_0}{4 \pi} i^2\left(\frac{d \vec{l} \times \vec{r}}{r}\right)$
C $d \vec{B}=\frac{\mu_0}{4 \pi} i^2\left(\frac{d \vec{l} \times \vec{r}}{r^2}\right)$
D $d \vec{B}=\frac{\mu_0}{4 \pi} i\left(\frac{d \vec{l} \times \vec{r}}{r^3}\right)$

The correct answer is:
$d \vec{B}=\frac{\mu_0}{4 \pi} i\left(\frac{d \vec{l} \times \vec{r}}{r^3}\right)$

Solution:
$d B=\frac{\mu_0}{4 \pi} \cdot \frac{i d l \sin \theta}{r^2} \Rightarrow d \vec{B}=\frac{\mu_0}{4 \pi} \cdot \frac{i(d \vec{l} \times \vec{r})}{r^3}$

Topic: Physics, Magnetic Effects of Current, JEE Main
The magnetic field due to a current carrying circular loop of radius $3 \mathrm{~cm}$ at a point on the axis at a distance of $4 \mathrm{~cm}$ from the centre is $54 \mu \mathrm{T}$. What will be its value at the centre of the Loop?

Options:
A
$250 \mu \mathrm{T}$

B
$150 \mu \mathrm{T}$

C
$125 \mu \mathrm{T}$

D
$75 \mu \mathrm{T}$

The correct answer is:
$250 \mu \mathrm{T}$

Solution:
Using formula $B=\frac{\mu_0^{\text {II }}}{2\left(R^2+X^2\right)^{3 / 2}}$, we get
$$
54=\frac{\mu_0 i(3)^2}{2\left[(3)^2+(4)^2\right]^{3 / 2}}
$$
At the centre of the coil, $X=0$ and $B=\frac{\mu_0 i}{2(3)}$
Using equation (i)
$$
B=\frac{54 \times 5^3}{(3)^2 \times 3} \Rightarrow B=250 \mu \mathrm{T} \text {. }
$$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2004
The magnetic field in a plane electromagnetic wave is $\mathrm{B}_{\mathrm{y}}=\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{T}$. The corresponding electric field will be :

Options:
A $E_y=10.5 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}$
B $E_z=1.17 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}$
C $E_y=1.17 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}$
D $E_z=105 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}$

The correct answer is:
$E_z=105 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}$

Solution:
$\begin{aligned}& \mathrm{E}_0=\mathrm{B}_0 \mathrm{C} \\& \mathrm{E}_0=3 \times 10^8 \times\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^3 \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right) \\& \mathrm{E}_0=105 \sin \left(1.5 \times 10^3 \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right) \mathrm{Vm}^{-1}
\end{aligned}$
Data inconsistent while calculating speed of wave. You can challenge for data.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The magnetic field in a plane electromagnetic wave is given by: By=2×10-7sinπ×103x+3π×1011t T. Calculate the wavelength.

Options:
A π×103 m
B 2×10-3 m
C 2×103 m
D π×10-3 m

The correct answer is:
2×10-3 m

Solution:
Recall the standard form of the equation of electromagnetic wave in terms of the magnetic field, By=B0sinkx+ωt T, here B0 is the magnitude of magnetic field, k=2πλ & ω=2πf. The equation is given in the question, By=2×10-7sinπ×103x+3π×1011t T

⇒λ=2πk=2ππ×103⇒λ=2×10-3 m.

Topic: Physics, Electromagnetic Waves, JEE Main
The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is :

Options:
A 9 V m-1
B 12 V m-1
C 3 V m-1
D 6 V m-1

The correct answer is:
6 V m-1

Solution:
E = B c=20×10-9×3×108
= 6 V/m

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2013 (07 Apr)
The magnetic field induction at the centre of a current-carrying circular coil (coil 1) and a closed coil (coil 2), shaped as a quarter of a disc is found to be equal in magnitude. If both the coils have equal area, then find the ratio of the currents flowing in coil 2 and coil 1.

Options:

The correct answer is:
8

Solution:
Let r and x be the radius of circular and quarter circular coil, then πr2=πx24 ⇒x=2r.
μ04π2πIr=μ04ππ2I'x ⇒I'I=4xr=8

Topic: Physics, Magnetic Effects of Current, JEE Main
The magnetic field inside a current carrying toroidal solenoid is $0.2 \mathrm{mT}$. What is the magnetic field inside the toroid if the current

Options:
A $0.02 \mathrm{mT}$
B $0.6 \mathrm{mT}$
C $0.8 \mathrm{mT}$
D $0.9 \mathrm{mT}$

The correct answer is:
$0.6 \mathrm{mT}$

Solution:
Magnetic field inside the toroid is given by
$$
\begin{aligned}
& \mathrm{B}=\mu_0 \mathrm{nI} \\
& \therefore \frac{\mathrm{B}_2}{\mathrm{~B}_1}=\frac{\mathrm{I}_2}{\mathrm{I}_1}=3
\end{aligned}
$$
(If cross-sectional radius of the solenoid is changed three will be no effect on the magnetic field, assuming the number of turns remains same.)
$$
\therefore \mathrm{B}_2=3 \mathrm{~B}_1=3 \times 0.2=0.6 \mathrm{~T}
$$

Topic: Physics, Magnetic Effects of Current, MHT CET, MHT CET 2021 (22 Sep Shift 2)
The magnetic field near a current carrying conductor is given by

Options:
A Coulomb's law
B Lenz' law
C Biot-savart's law
D Kirchoff's law

The correct answer is:
Biot-savart's law

Solution:
The formula is $B=\frac{\mu_o}{4 \pi} I \frac{\overrightarrow{d l} \times \vec{r}}{r^3}$
It is Biot-sarvart's law

Topic: Physics, Magnetic Effects of Current, JEE Main
The magnetic field of a given length of wire for a single turn coil at the centre is ' $B$ ' then, its value for two turns coil for the same wire is:

Options:
A $\frac{B}{4}$
B $\frac{B}{2}$
C $4 B$
D $2 B$

The correct answer is:
$4 B$

Solution:
Here $B_1=B=\frac{\mu_0 I}{2 R}$
$\begin{aligned}
& B_2=\frac{\mu_0(2 \mathrm{I})}{2 r} \\
& \therefore 2 \times 2 \pi r=2 \pi R \Rightarrow r=R / 2
\end{aligned}$
So, $B_2 = 4 B$

Topic: Physics, Magnetic Effects of Current, JEE Main
The magnetic field of a plane electromagnetic wave is B→=3×10-8sin200π(y+ct)i^ T . Where, c=3×108 m s-1 is the speed of light the corresponding electric filed is :

Options:
A E→=9sin[200π(y+ct)k^ V/m
B E→=-10-6sin[200π(y+ct)k^ V/m
C E→=3×10-8sin[200π(y+ct)k^ V/m
D E→=-9sin[200π(y+ct)k^ V/m

The correct answer is:
E→=9sin[200π(y+ct)k^ V/m

Solution:
E0=cB0

∴E→×B→‖C→⇒k^×i^=j^

E→=3×108×3×10-8sin[200πy+ct(-k^) =9sin[200πy+ct](-k^)

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The magnetic field of a plane electromagnetic wave is given by B→=2×10-8sin0.5×103x+1.5×1011tj^ T. The amplitude of the electric field would be

Options:
A 6Vm-1 along x-axis
B 3Vm-1 along z-axis
C 6Vm-1 along z-axis
D 2×10-8Vm-1 along z-axis

The correct answer is:
6Vm-1 along z-axis

Solution:
The ratio of electric field amplitude to the magnetic field amplitude of an electromagnetic wave is equal to the speed of light in the medium. Therefore,

c=E0B0⇒E0=cB0

⇒E0=3×108×2×10-8

⇒E0=6 V m-1

According to the given equation B→=2×10-8sin0.5×103x+1.5×1011tj^ T the wave travels in x direction. As electric field and magnetic field vibrate perpendicular to the velocity of light, electric field can not be along x axis.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2022 (26 Jul Shift 1)
The magnetic field of a plane electromagnetic wave is given by $\overrightarrow{\mathrm{B}}=3 \times 10^{-8} \cos \left(1.6 \times 10^3 x+\right.$ $48 \times 10^{10}$ t) $\hat{j}$, then the associated electric field will be:

Options:
A $3 \times 10^{-8} \cos \left(1.6 \times 10^3 x+48 \times 10^{10} t\right) \hat{i} / \mathrm{m}$
B $3 \times 10^{-8} \sin \left(1.6 \times 10^3 x+48 \times 10^{10} t\right) \hat{i} / \mathrm{m}$
C $9 \sin \left(1.6 \times 10^3 x-48 \times 10^{10} t\right) \hat{k} \mathrm{~V} / \mathrm{m}$
D $9 \cos \left(1.6 \times 10^3 x+48 \times 10^{10} t\right) \hat{k} \mathrm{~V} / \mathrm{m}$

The correct answer is:
$9 \cos \left(1.6 \times 10^3 x+48 \times 10^{10} t\right) \hat{k} \mathrm{~V} / \mathrm{m}$

Solution:
Given, $\mathrm{B}=3 \times 10^{-8} \cos \left(1.6 \times 10^3 x\right.$ $\left.+48 \times 10^{10} t\right)$
On comparing equations (1) & (2), we get;
$\begin{aligned}
\mathrm{B}_0 & =3 \times 10^{-8} \\
c & =\frac{\omega}{k}=\frac{48 \times 10^{10}}{1.6 \times 10^3} \\
& =3 \times 10^8
\end{aligned}$
And also, $c=\frac{\mathrm{E}_0}{\mathrm{~B}_0}$
Hence, $E_0=B_0 \times c$
$\begin{aligned}
& =3 \times 10^{-8} \times 3 \times 10^8 \\
& =9
\end{aligned}$
So, the required equation is:
$\mathrm{E}=9 \cos \left(1.6 \times 10^3 x+48 \times 10^{10} t\right)$

Topic: Physics, Electromagnetic Waves, JEE Main
The magnetic field of a plane electromagnetic wave is given by B→=B0i^coskz-ωt+B1j^coskz+ωt, where B0=3×10–5 T and B1=2×10–6 T. The RMS value of the force experienced by a stationary charge Q=10–4 C at z=0 is closest to

Options:
A 0.1 N
B 0.9 N
C 3×10–2 N
D 0.6 N

The correct answer is:
0.6 N

Solution:
B→=B0i^coskz-ωt+B1j^coskz+ωt
B0=3×10-5 T & B1=2×10-6 T
Electric filed associated with it is,
E→=-B0cj^coskz-ωt-B1ci^coskz+ωt
Here, c is the speed of light in vacuum.
At z=0,
Frms=QB0c2+QB1c22
Here c is speed of light in vacuum.
=10-4×3×10-5×3×1082+10-4×2×10-6×3×10822
=0.81+0.00362=0.6 N

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The magnetic field of a plane electromagnetic wave is given by Bâ=B0i^coskz-Ït+B1j^coskz+Ït, where B0=3Ã10â5 T and B1=2Ã10â6 T. The RMS value of the force experienced by a stationary charge Q=10â4 C at z=0 is closest to

Options:
A 0.1 N
B 0.9 N
C 3×10–2 N
D 0.6 N

The correct answer is:
0.6 N

Solution:
B→=B0i^coskz-ωt+B1j^coskz+ωt
B0=3×10-5 T B1=2×10-6 T
Electric filed associated with it is,
E→=-B0cj^coskz-ωt-B1ci^coskz+ωt
Here, c is speed of light in vacuum.
At z=0,
Frms=QB0c2+QB1c22
Here c is speed of light in vacuum.
=10-4×3×10-5×3×1082+10-4×2×10-6×3×10822
=0.81+0.00362=0.6 N

Topic: Physics, Electromagnetic Waves, JEE Main