datavorous/entrance-exam-dataset huggingface dataset:
A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature $\theta$ along the length $x$ of the bar from its hot end is best described by which of the following figure.

Options:
A

B

C

D

The correct answer is:

Solution:
We know that $\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{KA} \frac{\mathrm{d} \theta}{\mathrm{dx}}$
In steady state flow of heat
$$
\begin{aligned}
& \mathrm{d} \theta=\frac{\mathrm{dQ}}{\mathrm{dt}} \cdot \frac{1}{\mathrm{kA}} \cdot \mathrm{dx} \\
& \Rightarrow \theta_{\mathrm{H}}-\theta=\mathrm{k}^{\prime} \mathrm{x} \Rightarrow \theta=\theta_{\mathrm{H}}-\mathrm{k}^{\prime} \mathrm{x}
\end{aligned}
$$
Equation $\theta=\theta_H-k^{\prime} x$ represents a straight line.

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2009
A long, rectangular conducting loop of width $l$ mass $m$ and resistance $R$ is placed partly in a perpendicular magnetic field $B$. It is pushed downwards with velocity $v$ so that it may continue to fall freely. The velocity $v$ is $(g=$ acceleration due to gravity)

Options:
A $\frac{m g R^2}{B l}$
B $\frac{m g R}{B^2 l^2}$
C $\frac{m g l}{B^2 R^2}$
D $\frac{B^2 l^2 R}{m g}$

The correct answer is:
$\frac{m g R}{B^2 l^2}$

Solution:
The motional emf is given by: $V=B v l$
Current in the loop,
$i=\frac{V}{R}=\frac{B v l}{R}$
During free fall the electromagnetic force felt by the loop balances its weight, and since the loop is long so we assume its falling with a terminal velocity.
$\begin{aligned}
& \therefore i l B=m g \\
& \Rightarrow B\left(\frac{B v l}{R}\right) l=m g \\
& \Rightarrow v=\frac{m g R}{B^2 l^2}
\end{aligned}$

Topic: Physics, Electromagnetic Induction, JEE Main
A long rigid wire lies along the X- axis and carries a current of 10 A in the positive X- direction. Round the wire, the external magnetic field is Bâ=i^+2x2j^ with x in meters and B in Tesla. The magnetic force (in SI units) on the segment of the wire between x=1 m and x=4 m is

Options:
A 1260
B 12803
C 1310
D 420

The correct answer is:
420

Solution:
dFâ=IdlâÃBâ
F=Iâ«dxi^Ãi^+2x2j^
=10 â«x=1x=42x2 dx k^
=20x33x=1x=4=20343-13=420 N
dFââ=âIâ(dlââÃBâ)
F=Iâ«(dxâi^)âÃâ(i^+2x2âj^)
=10ââ«x=1x=42x2âdxâk^
=20â[x33]x=1x=4=203[43-13]=420âN

Topic: Physics, Magnetic Effects of Current, JEE Main
A long silver teaspoon is placed in a cup filled with hot tea. After some time, the exposed end (the end which is not dipped in tea) of the spoon becomes hot even without direct contact with the tea. This phenomenon can be explained mainly by

Options:
A Conduction
B Reflection
C Radiation
D Thermal expansion

The correct answer is:
Conduction

Solution:
As atoms in the spoon vibrates about their equilibrium positions and transfer energy form one end to other end. This process is called conduction.

Topic: Physics, Thermal Properties of Matter, NEET
A long solenoid carrying a current produces a magnetic field $B$ along its axis. If the current is doubled and the number of turns per $\mathrm{cm}$ is halved, the new value of the magnetic field is:

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

The correct answer is:
$B$

Solution:
Magnetic field induction at point insides the solenoid of length $l$, having $n$ turns per unit length carrying current $i$. is given by
$$
B=\mu_0 n i
$$
If $i$ is double and $n$ is halved then $B$ remains same.

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2003
A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of magnetic field will be equal to

Options:
A 2B
B 4B
C B2
D B

The correct answer is:
B

Solution:
Magnetic field due to long solenoid along its axis is given as,

B=μ0nI, where n is the number of turns in the solenoid per unit length, I is current in the solenoid and μ0 is a constant called the permeability of free space.

⇒B1 B2=n1I1n2I2

According to the question, n2=n12 and I2=2I1

⇒BB2=n1n12×I12I1⇒B2=B; means, the magnetic field remains the same.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (25 Jun Shift 2)
A long solenoid has 20 turns per cm. A small loop of area 4π cm2 is placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 1.0 A to 3.0 A in 0.2 s, what is the magnitude of the induced emf in the loop while the current is changing?

Options:
A 2.4 μV
B 3.2 μV
C 7.2 μV
D 4.8 μV

The correct answer is:
3.2 μV

Solution:
Magnetic field due to solenoid along its axis,

B=μ0nI=4π×10-720×102I=8π×10-4I T

Now induced EMF due to change in magnetic flux can be written as,

ε=∆ϕ∆t=B2A-B1A0.2=5A8π×10-43-1=5×4π×10-416π×10-4=3.2 μV

Topic: Physics, Electromagnetic Induction, JEE Main
A long solenoid has 200 turns per $\mathrm{cm}$ and carries a current $\mathrm{i}$. The magnetic field at its centre is $6.28 \times 10^{-2}$ Weber $/ \mathrm{m}^2$. Another long solenoid has 100 turns per $\mathrm{cm}$ and it carries a current $\mathrm{i} / 3$. The value of the magnetic field at its centre is

Options:
A
$1.05 \times 10^{-4}$ Weber/m

B
$1.05 \times 10^{-2}$ Weber $/ \mathrm{m}^2$

C
$1.05 \times 10^{-5} \mathrm{Weber} / \mathrm{m}^2$

D
$1.05 \times 10^{-3} \mathrm{Weber} / \mathrm{m}^2$

The correct answer is:
$1.05 \times 10^{-2}$ Weber $/ \mathrm{m}^2$

Solution:
$B_2=\frac{B_1 n_2 i_2}{n_1 i_1}=\frac{\left(6.28 \times 10^{-2}\right)(100 \times i / 3)}{200(i)}=1.05 \times 10^{-2} \mathrm{~W} / \mathrm{m}^2$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2006
A long solenoid has 200 turns per $\mathrm{cm}$ and carries a current of $2.5 \mathrm{amps}$. The magnetic field at its centre is $\left(\mu_0=4 \pi \times 10^{-7}\right.$ weber /amp $\left.-m\right)$

Options:
A $3.14 \times 10^{-2}$ weber $/ \mathrm{m}^2$
B $6.28 \times 10^{-2}$ weber $/ \mathrm{m}^2$
C $9.42 \times 10^{-2}$ weber $/ \mathrm{m}^2$
D $12.56 \times 10^{-2}$ weber $/ \mathrm{m}^2$

The correct answer is:
$6.28 \times 10^{-2}$ weber $/ \mathrm{m}^2$

Solution:
$B=\mu_0 n i=4 \pi \times 10^{-7} \times \frac{200}{10^{-2}} \times 2.5=6.28 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^2$

Topic: Physics, Magnetic Effects of Current, JEE Main
A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} \omega \mathrm{b}$. The selfinductance of the solenoid is

Options:
A 4.0 henry
B 2.5 henry
C 2.0 henry
D 1.0 henry

The correct answer is:
1.0 henry

Solution:
$$
\begin{aligned}
& N \phi=L i \\
& 500 \times 4 \times 10^{-3}=2 L \\
& L=1.0 \text { henry }
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, NEET, NEET 2008 (Mains)
A long solenoid has 1000 turns. When a current of 4 A flows through it the magnetic flux linked with each turn of the solenoid is 4×10-3 Wb. The self-inductance of the solenoid is:

Options:
A 4  H
B 3  H
C 2  H
D 1  H

The correct answer is:
1  H

Solution:
In the case of self-inductance of the solenoid, magnetic flux ϕ is directly proportional to the current i flowing in the coil. The flux through each turn is equal to the flux linkage.

The relation is given as Nϕ=Li, here N is the number of turns and L is inductance.

So putting the given values, we get

1000×4×10-3=L×4

Thus, the self inductance is 1 H.

Topic: Physics, Electromagnetic Induction, JEE Main
A long solenoid having 100 turns per $\mathrm{cm}$ carries a current of $\frac{4}{\pi}$ A. At the centre of it is placed a coil of 200 turns of cross-sectional area $25 \mathrm{~cm}^2$ having its axis parallel to the field produced by the solenoid. When the direction of the current in the solenoid is reversed with in $0.04 \mathrm{~s}$, the induced emf in the coil is

Options:
A $0.2 \mathrm{~V}$
B $0.4 \mathrm{~V}$
C $0.002 \mathrm{~V}$
D $0.016 \mathrm{~V}$

The correct answer is:
$0.4 \mathrm{~V}$

Solution:
Due to solenoid a magnetic field exists around coil placed at centre
Magnetic field intensity produced by solenoid,
$B=\mu_0 n I$
Flux linked with coil is
$\phi_B=N . B \cdot A$
Where, $N=$ Number of turns in coil
$A=$ Area of coil
As current in solenoid is reversed, change in flux is
$\Delta \phi_B=2 N \cdot B \cdot A$
This change in flux produces an emf in coil given by
Induced emf $e=$ Rate of change of flux
$\Rightarrow \quad e=\frac{2 N B A}{\Delta t}=\frac{2 N\left(\mu_0 n I\right) A}{\Delta t}$
Here $N=200, A=25 \mathrm{~cm}^2=25 \times 10^{-4} \mathrm{~m}^2$
$I=\frac{4}{\pi} \mathrm{A}, n=100$ tuns $/ \mathrm{cm}=100 \times 100$ turns $/ \mathrm{m}$
$\mu_0=4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{mA}^{-1}$ and $\Delta t=0.04 \mathrm{~s}$
So induced emf,
$e=\frac{2 \times 200 \times 4 \pi \times 10^{-7} \times 10^4 \times \frac{4}{\pi} \times 25 \times 10^{-4}}{0.04}=0.4 \mathrm{~V}$
Induced emf in coil is $0.4 \mathrm{~V}$

Topic: Physics, Electromagnetic Induction, JEE Main
A long solenoid is carrying a current $I=I_0 \sin (\omega t)$, having $N$ turns per unit length and radius $R$. A square loop is placed inside the solenoid with its plane perpendicular to the solenoid axis, and its corners touching the solenoid. Now the emf induced in the square coil.

Options:
A $\mu_0 N I_0 R^2 \sin (\omega t)$
B $2 \mu_0 N l_0 R^2 \sin (\omega t)$
C $2 \mu_0 N I_0 \dot{R}^2 \omega \cos (\omega t)$
D $\mu_0 N I_0 R^2 \pi \omega \cos (\omega t)$

The correct answer is:
$2 \mu_0 N I_0 \dot{R}^2 \omega \cos (\omega t)$

Solution:
Given, $I=I_0 \sin \omega t$
Magnetic field inside the solenoid, $B=\mu_0 N I$
$$
\Rightarrow \quad B=\mu_0 N I_0 \sin \omega t
$$
When side of square loop be $l$, then
$$
\begin{array}{rlrl}
& & 4 l & =2 \pi R \\
\Rightarrow \quad & l & =\frac{\pi R}{2}
\end{array}
$$
Area of square loop, $A=l^2$
$$
=\left(\frac{\pi R}{2}\right)^2=\frac{\pi^2 R^2}{4}
$$
Magnetic flux linked with square loop,
$$
\begin{aligned}
\phi & =B A \\
& =\mu_0 N I_0 \sin \omega t \times \frac{\pi^2 R^2}{4} \\
\phi & =\frac{\mu_0 \pi^2 R^2 N I_0 \sin \omega t}{4}
\end{aligned}
$$
Induced emf in the square coil, $e=\frac{d \phi}{d t}$
$$
\begin{aligned}
& =\frac{d}{d t} \frac{\mu_0 \pi^2 R^2 N I_0 \sin \omega t}{4} \\
& =\frac{\mu_0 \pi^2 R^2 N I_0 \omega \cos \omega t}{4} \quad\left(\because \frac{\pi^2}{4}=246\right) \\
& =2,46 \mu_0 R^2 N I_0 \omega \cos \omega t
\end{aligned}
$$
Which is closest to the value given in option (c).

Topic: Physics, Electromagnetic Induction, JEE Main
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 ______________.

(µ0= 4π× 10–7 T m A–1)

Options:
A 1232×10–4 T
B 176×10–4 T
C 352×10–4 T
D 88×10–4 T

The correct answer is:
176×10–4 T

Solution:
Magnetic field inside solenoid =μoni

where n= Number of turns per unit length

n=7000 turns/m

Bsolenoid =μoni=4π×10-7×7000×2

=8×227×10-6×7×100

=176×10-4 T

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (24 Jan Shift 2)
A long solenoid of cross-sectional radius R has a thin insulated wire ring of uniform cross-section tightly put on its winding. The ring is made out of two different material such that one half of the ring has a resistance 10 times that of the other half. The magnetic induction produced by the solenoid varies with time as B=bt, where b is a constant. The magnitude of the electric field strength in the ring (other than the induced electric field) is

Options:
A 911Rb
B 922Rb
C 9 Rb
D Rb

The correct answer is:
922Rb

Solution:
Let us assume that the current in the ring is i and the resistance of the two halves are r and 10r, then by faraday's law

dÏdt=ÏR2b=ir+i10r

âi=ÏR2b11r

The induced electric field in both the parts of the ring is

Eind=R2dBdt=Rb2

Let E be the field which is developed in each part due to charge accumulation at the junctions, then for the upper half

Eind+EÏR=i10r

and for the lower half

Eind-EÏR=ir
â2EÏR-9ir=0

E=9ir2ÏR

âE=9r2ÏRÃÏR2b11r=922Rb

Topic: Physics, Electromagnetic Induction, JEE Main
A long solenoid of diameter 0.1 m has 2×104 turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 10π2 Ω, the total charge flowing through the coil during this time is

Options:
A 32Ï Î¼C
B 16Â Î¼C
C 32 Î¼C
D 16ÏÂ Î¼C

The correct answer is:
32 Î¼C

Solution:
In the given case, the solenoid produces a changing magnetic field. This field is linked with the coil inside and hence, there is an induced emf in the coil.

We have, from the laws of EMI:

ε= -Ndϕdt
εR=NRdϕdt

The ratio given above is the induced current in the circuit which can further be written as the rate of flow of charge per unit time.
dq=NRdϕ
ΔQ=NΔϕR
ΔQ=ΔϕtotalR
=NBAR
= N μ 0 niπ r 2 R
Putting values
=100×4π×10−7×2×104×4×π×(0.01)210π2
ΔQ=32 μC

Topic: Physics, Electromagnetic Induction, JEE Main
A long solenoid of radius R carries a time t dependent current It=I0t1-t . A ring of radius 2R is placed coaxially near its middle. During the time interval 0â¤tâ¤1, the induced current IR and the induced EMFVR in the ring change as:

Options:
A Direction of IR remains unchanged and VR is maximum at t=0.5
B At t=0.25 direction of IR reverses and VR is maximum
C Direction of IR remains unchanged and VR is zero at t=0.25
D At t=0.5 direction of IR reverses and VR is zero

The correct answer is:
At t=0.5 direction of IR reverses and VR is zero

Solution:
I=I0t-I0t2
ϕ=BA
ϕ=μ0nIA
VR=-dϕdt=-μ0nAI01-2t
VR=0 at t=12
And IR=VRResistance of loop=0

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (07 Jan Shift 1)
A long solenoid of radius R carries a time t dependent current It=I0t1-t . A ring of radius 2R is placed coaxially near its middle. During the time interval 0â¤tâ¤1, the induced current IR and the induced EMFVR in the ring change as:

Options:
A Direction of IR remains unchanged and VR is maximum at t=0.5
B At t=0.25 direction of IR reverses and VR is maximum
C Direction of IR remains unchanged and VR is zero at t=0.25
D At t=0.5 direction of IR reverses and VR is zero

The correct answer is:
At t=0.5 direction of IR reverses and VR is zero

Solution:
I=I0t-I0t2
Ï=BA
Ï=Î¼0nIA
VR=-dÏdt=-Î¼0nAI01-2t
VR=0 at t=12
And IR=VRResistance of loop

Topic: Physics, Electromagnetic Induction, NEET
A long solenoid of the cross-sectional radius R has a thin insulated wire ring tightly put on its winding. One half of the ring has the resistance 10 times that of the other half. The magnetic induction produced by the solenoid varies with time as B = bt, whose b is a constant. If the magnitude of the electric field strength in the ring is pqRb, where p and q are the smallest positive integers, then what is the value of p+q?

Options:

The correct answer is:
31

Solution:
Both upper half and lower half will have same effective area of Ï Râ¡ 2 2 so charge in flux will be same and induced emf will have some value. But since the resistance is different due to which current must be different but ring is as a whole is closed circuit so electric field will be generated to make the current flow in both parts to be same.
Îµ + Eâ¡ Ï Râ¡ - iâ¡ 1 0 râ¡ = 0 â¦ iâ¡
Îµ - Eâ¡ Ï Râ¡ - iâ¡râ¡ = 0 â¦ iâ¡ iâ¡
â 2 E Ï Râ¡ - 9 iâ¡ râ¡ = 0
Eâ¡ = 9 i r 2 Ï Râ¡ â iâ¡ = Eâ¡ 1 1 râ¡ = Ï Râ¡ 2 b 1 1 râ¡
â Eâ¡ = 9 r 2 Ï Râ¡ Ã Ï Râ¡ 2 b 1 1 râ¡ = 9 2 2 Rb
â´ 9 2 2 Rb

Topic: Physics, Electromagnetic Induction, JEE Main
A long spring is stretched by $2 \mathrm{~cm}$ and its potential energy is $U$. If the spring is stretched by $10 \mathrm{~cm}$; its potential energy will be

Options:
A $U / 5$
B $U / 25$
C $5 U$
D $25 U$

The correct answer is:
$25 U$

Solution:
The potential energy of a stretched spring is
$$
U=\frac{1}{2} k x^{2}
$$
Here, $k=$ spring constant, $x=$ elongation in spring. But given that, the elongation is $2 \mathrm{~cm}$.
So, $U=\frac{1}{2} k(2)^{2}$
$$
\Rightarrow \quad U=\frac{1}{2} k \times 4 ...(i)
$$
If elongation is $10 \mathrm{~cm}$ then potential energy
$$
\begin{aligned}
U^{\prime} &=\frac{1}{2} k(10)^{2} \\
U^{\prime} &=\frac{1}{2} k \times 100 ...(ii)
\end{aligned}
$$
On dividing Eq. (ii) by Eq. (i), we have
$$
\frac{U^{\prime}}{U}=\frac{\frac{1}{2} k \times 100}{\frac{1}{2} k \times 4}
$$
or $\frac{U^{\prime}}{U}=25 \Rightarrow U^{\prime}=25 U$

Topic: Physics, Work Power Energy, JEE Main
A long spring, when stretched by a distance $x$, has potential energy $U$. On increasing the stretching to $n x$, the potential energy of the spring will be

Options:
A $\frac{U}{n}$
B $n U$
C $n^2 U$
D $\frac{U}{n^2}$

The correct answer is:
$n^2 U$

Solution:
Potential energy of the spring,
and
$U^{\prime}=\frac{1}{2} k(n x)^2 \Rightarrow U^{\prime}=n^2 \frac{1}{2} k x^2$
$\Rightarrow \quad U^{\prime}=n^2 U \quad \text{[From Eq. (i)]}$

Topic: Physics, Work Power Energy, JEE Main
A long straight wire carrying a current of $30 \mathrm{~A}$ is placed in an external uniform magnetic field of induction $4 \times 10^{-4} \mathrm{~T}$. The magnetic field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point $2.0 \mathrm{~cm}$ away from the wire is
$\left(\mu_0=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\right)$

Options:
A (a) $10^{-4}$
B $3 \times 10^{-4}$
C $5 \times 10^{-4}$
D $6 \times 10^{-4}$

The correct answer is:
$5 \times 10^{-4}$

Solution:
$i=30 \mathrm{~A}, B=4 \times 10^{-4} \mathrm{~T}$
$r=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}$
Initial magnetic field (parallel to the wire),
$B_1=4 \times 10^{-4} \mathrm{~T}$
Magnetic field produced by the straight wire,
$B_2=\frac{\mu_0 i}{2 \pi r}=\frac{2 \times 10^{-7} \times 30}{2 \times 10^{-2}}$
$=3 \times 10^{-4} \mathrm{~T}$
$B_2$ will be in the plane perpendicular to the plane of wire, so $B_1$ and $B_2$ are perpendicular to each other.
$\therefore$ Resultant magnetic field,
$B=\sqrt{B_1^2+B_2^2}$
$=\sqrt{\left(4 \times 10^{-4}\right)^2+\left(3 \times 10^{-4}\right)^2}$
$=5 \times 10^{-4}$

Topic: Physics, Magnetic Effects of Current, JEE Main
A long straight wire carrying a current of 30 A is placed in an external uniform magnetic field of induction 4×10–4 T . The direction of external magnetic field is along the direction of the current. The magnitude of the resultant magnetic induction at a point 2.0 cm away from the wire is n×10-4 T. What is the value of n?

Options:

The correct answer is:
5

Solution:
Magnetic field due to wire
B=Î¼0I2Ïr=4ÏÃ10-72ÏÃ302Ã10-2
=3Ã10-4 T
This magnetic field will be perpendicular to external magnetic field.
â´ Net magnetic field
B=B2+B02
=3Ã 10â4 2+4Ã10â4 2
=5Ã10â4 T

Topic: Physics, Magnetic Effects of Current, JEE Main
A long straight wire carrying current $I$ and a rectangular frame with side lengths $a$ and $b$ lie in the same plane as shown in the figure. The mutual inductance of the wire and frame is

Options:
A $\frac{\mu_0}{2 \pi} a b$
B $\frac{\mu}{4 \pi} \frac{b}{a}$
C $\frac{\mu_0 a}{2 \pi} \ln 2$
D $\frac{\mu_0 b}{2 \pi} \ln 2$

The correct answer is:
$\frac{\mu_0 b}{2 \pi} \ln 2$

Solution:
Magnetic field through the wire, $B=\frac{\mu_0 I b}{2 \pi a}$.
$\begin{array}{ll}\therefore \text { Magnetic flux induced, } \phi & =\oint_s \mathbf{B} \cdot d \mathbf{a} \\ \Rightarrow & \phi=\frac{\mu_0 I b}{2 \pi} \int_a^{a+a} \frac{d a}{a} \\ \Rightarrow & \phi=\frac{\mu_0 I b}{2 \pi} \ln 2\end{array}$
$\therefore$ Mutual inductance of the wire and the frame is
$$
M=\frac{\phi}{I}=\frac{\mu_0 b}{2 \pi} \ln 2
$$

Topic: Physics, Electromagnetic Induction, JEE Main
A long straight wire of radius ' $a$ ' caries a steady current $\mathrm{i}$. The current is uniformly distributed across its cross section. The ratio of the magnetic field at $\frac{a}{2}$ and $2 a$ is

Options:
A
$\frac{1}{4}$

B
$4$

C
$1$

D
$\frac{1}{2}$

The correct answer is:
$1$

Solution:
$\mathrm{B} 2 \pi \frac{\mathrm{a}}{2}=\mu_0 \frac{\mathrm{i}}{\pi \mathrm{a}^2}\left(\frac{\pi \mathrm{a}^2}{4}\right)$
$\mathrm{B}_1=\frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{a}} \quad \ldots$ (i)
$\mathrm{B}_2 2 \pi(2 \mathrm{a})=\mu_0 \mathrm{i}$
$\mathrm{B}_2=\frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{a}} \quad \dots (ii)$
$\frac{\mathrm{B}_1}{\mathrm{~B}_2}=1$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2007
A long, straight Wire of radius a carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance a3 and 2a , respectively from the axis of the wire is:

Options:
A 23
B 2
C 12
D 32

The correct answer is:
23

Solution:
BA=μ0ir2πa2=μ0ia32πa2=μ0iπa2a6=μ0i6πa
BB=μ0i2π2a
BABB=46=23

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (09 Jan Shift 1)
A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of the magnetic field at $\frac{a}{2}$ and $2 a$ from axis of the wire is :

Options:
A $1: 4$
B $1: 1$
C $3: 4$
D $4: 1$

The correct answer is:
$1: 1$

Solution:
$\begin{aligned} & \mathrm{B}_1 2 \pi \frac{\mathrm{a}}{2}=\mu_{\mathrm{o}} \frac{\mathrm{I}}{4} \\ & \mathrm{~B}_1=\frac{\mu_{\mathrm{o}} \mathrm{I}}{4 \pi \mathrm{a}} \\ & \mathrm{B}_2 2 \pi 2 \mathrm{a}=\mu_{\mathrm{o}} \mathrm{I} \\ & \mathrm{B}_2=\frac{\mu_{\mathrm{o}} \mathrm{I}}{4 \pi \mathrm{a}}\end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (08 Apr Shift 2)
A long straight wire of radius \( a \) carries a steady current \( i \). The current is uniformly distributed across its cross section. The ratio of the magnetic field at \( a / 2 \) and \( 2 a \) is,

Options:
A \( 1 / 2 \)
B \( 1 / 4 \)
C \( 4 \)
D \( 1 \)

The correct answer is:
\( 1 \)

Solution:
Uniform current is flowing. Current enclosed in the first amperean path is,

I · π r 1 2 π R 2 = I r 1 2 R 2
∴ B = μ 0 × current path = μ 0 · I r 1 2 2 π r 1 R 2 = μ 0 I r 1 2 π R 2
Magnetic induction at a distance,

r 2 = μ 0 · I 2 π r 2
∴ B 1 B 2 = r 1 r 2 R 2 = a 2 · 2 a a 2 = 1.

Topic: Physics, Magnetic Effects of Current, JEE Main
A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance rr<R from its centre will be

Options:
A B∝r
B B∝1r
C B∝r2
D B∝1r2

The correct answer is:
B∝r

Solution:
Using Ampere's law and noting that the steady current is uniformly distributed over the cross-section of the cylindrical wire.

B=μ0Ir2πR2, r≤Rμ0I2πr, r>R

Thus, magnetic field B∝r from the centre for r<R.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (25 Jun Shift 1)
A long string is stretched by $2 \mathrm{~cm}$ and the potential energy is $\mathrm{V}$. If the spring is stretched by $10 \mathrm{~cm}$, its potential energy will be

Options:
A $\mathrm{V} / 25$
B $\mathrm{V} / 5$
C $5 \mathrm{~V}$
D $25 \mathrm{~V}$

The correct answer is:
$25 \mathrm{~V}$

Solution:
$\quad V=\frac{1}{2} k(x)^{2}=\frac{1}{2} k(2)^{2}$ or $k=\frac{2 V}{4}=\frac{V}{2}$
$$
\mathrm{V}^{\prime}=\frac{1}{2} \mathrm{k}(10)^{2}=\frac{1}{2} \times\left(\frac{\mathrm{V}}{2}\right)(10)^{2}=25 \mathrm{~V}
$$

Topic: Physics, Work Power Energy, JEE Main
A long thin magnet of magnetic moment M is bent into a semi-circle. The decrease in the magnetic moment is

Options:
A 2Mπ
B πM2
C Mπ-2π
D M2-π2

The correct answer is:
Mπ-2π

Solution:
Magnetic length is the minimum distance between the north pole and the south pole, which is decreased in the second case.
M′=2Mπ, decrease in magnetic moment
=M-2Mπ

Topic: Physics, Magnetic Properties of Matter, JEE Main
A long thin non-conducting cylindrical pipe of radius R carrying a uniform surface charge density Ï is rotating about its axis with a constant angular speed Ï. The energy density of the magnetic field inside the tube is

Options:
A μ0σ2ω2R42
B μ0σ2ω2R22
C μ0σ2ω2R24
D μ0σ2ω2R44

The correct answer is:
μ0σ2ω2R22

Solution:
The rotating cylinder is equivalent to a solenoid with current per unit length =σωR

∴ B=μ0σωR

∴ u=12μ0B2=12μ0μ02· σωR2

u= μ0σ2ω2R22

Topic: Physics, Magnetic Effects of Current, JEE Main