datavorous/entrance-exam-dataset huggingface dataset:
A long hollow copper tube carries a current I. Then which of the following will be true?

Options:
A The magnetic field B will be zero at all points inside the tube
B The magnetic field B will be zero only at points on the axis of the tube
C The magnetic field B will be maximum at points on the axis of the tube
D The magnetic field will be zero at any point outside the tube

The correct answer is:
The magnetic field will be zero at any point outside the tube

Solution:
The magnetic field B will be uniform inside the hollow tube, excepts near the ends. Also magnetic field is zero at any point outside the tube.

Topic: Physics, Magnetic Effects of Current, NEET
A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature θ 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 dQdt=kAdθdx
In steady-state flow of heat
dθ= dQ dt . 1 kA dx
⇒ θH-θ=k′x
⇒ θ=θH-k′x
Equation θ=θH-k′x represents a straight line.

Topic: Physics, Thermal Properties of Matter, NEET
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, 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 cm is halved, then the new value of the magnetic field is

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

The correct answer is:
B

Solution:
Magnetic field on the axis of a solenoid
B=μ0ni
B′=μ0n22i=B

Topic: Physics, Magnetic Effects of Current, NEET
A long solenoid has 500 turns. When a current of $2 \mathrm{~A}$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} \mathrm{~Wb}$. The self-inductance of the solenoid is

Options:
A $2.5 \mathrm{H}$
B $2.0 \mathrm{H}$
C $1.0 \mathrm{H}$
D $4.0 \mathrm{H}$

The correct answer is:
$1.0 \mathrm{H}$

Solution:
Key Idea : Inductance of a coil is numerically equal to the emf induced in the coil when the current in the coil changes at the rate of $1 \mathrm{As}^{-1}$. If $I$ is the current flowing in the circuit, then flux linked with the circuit is observed to be proportional to $I$, ie,
$\begin{aligned}
& \phi \propto I \\
& \text{or } \phi=L I
\end{aligned}$
where $L$ is called the self-inductance or coefficient of self-inductance or simply inductance of the coil.
Net flux through solenoid,
$\phi=500 \times 4 \times 10^{-3}=2 \mathrm{~Wb}$
or $2=L \times 2$ [after putting values in Eq. or $L=1 \mathrm{H}$

Topic: Physics, Electromagnetic Induction, NEET, NEET 2008 (Screening)
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, NEET, NEET 2016 (Phase 1)
A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is:
μ0=4π×10−7T m A−1

Options:
A 3.14×10–4 T
B 6.28×10–5 T
C 3.14×10–5 T
D 6.28×10–4 T

The correct answer is:
6.28×10–4 T

Solution:
The magnetic field at the centre of the solenoid is given by, B=μ0ni

Where n is the number of turns per unit length
⇒B=4π×10-7×100(0.5)×2.5
⇒B=6.28×10-4 T

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2020 (Phase 1)
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, NEET, NEET 2017
A long solenoid of length L has a mean diameter D. It has n layers of winding of N turns each. If it carries a current I, the magnetic field at its centre will be

Options:
A Proportional to D
B Inversely proportional to D
C Independent of D
D Proportional to L

The correct answer is:
Independent of D

Solution:
Magnetic field due to solenoid is independent of diameter
∵B=μ0 nI

Topic: Physics, Magnetic Effects of Current, NEET
A long solenoid of radius 1 mm has 100 turns per mm. If 1 A current flows in the solenoid, the magnetic field strength at the centre of the solenoid is:

Options:
A 12.56×10-2 T
B 12.56×10-4 T
C 6.28×10-4 T
D 6.28×10-2 T

The correct answer is:
12.56×10-2 T

Solution:
Radius =1 mm

N=100 turns per mm

Current =1 A

Magnetic field strength at the centre of the solenoid is,

B=μ0ni=4π×10-7×100×103×1⇒B=12.56×10-2 T

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2022 (Phase 1)
A long solenoid with 40 turns per cm carries a current of 1 A. The magnetic energy stored per unit volume is ....... J m-3.

Options:
A 3.2 Ï
B 32Â Ï
C 1.6Â Ï
D 6.4Â Ï

The correct answer is:
3.2 Ï

Solution:
n=40 turns cm-1
=40Ã102 turns m-1
I=1.00 A
The magnetic energy stored per unit volume in a solenoid
U=B22Î¼0,B=Î¼0nI
(Magnetic field inside the solenoid)
â´ U=B22Î¼0=Î¼0nI22Î¼0
=Î¼02n2I22Î¼0=Î¼0n2I22
=4ÏÃ10-7Ã40Ã1022Ã122
=3.2 Ï J m-3

Topic: Physics, Magnetic Effects of Current, NEET
A long spring is stretched by 3 cm and its potential energy is V. If the spring is stretched by 6 cm, its potential energy will be

Options:
A V
B 2V
C 3V
D 4V

The correct answer is:
4V

Solution:
P.E.=12×Y×strain2=12×Y×l2L2
or P.E P.E.∝l2
∴ P.E2P.E1= l22l12= 6232= 369=4
∴ P.E.2=4 P.E1=4 V

Topic: Physics, Work Power Energy, NEET
A long spring is stretched by x cm its P.E. is U. If the same spring is stretched by Nx cm the P.E. stored in it will become

Options:
A U/N
B NU
C N2U
D U/N3

The correct answer is:
N2U

Solution:
U=12Kx2

U′=12KNx2=12Kx2N2=UN2

Topic: Physics, Work Power Energy, NEET
A long straight wire carries a certain current and produces a magnetic field $2 \times 10^{-4} \frac{\text { Weber }}{\mathrm{m}^2}$ at a perpendicular distance of $5 \mathrm{~cm}$ from the wire. An electron situated at $5 \mathrm{~cm}$ from the wire moves with a velocity $10^7 \mathrm{~m} / \mathrm{s}$ towards the wire along perpendicular to it. The force experienced by the electron will be (change on electron $1.6 \times 10^{-19} \mathrm{C}$ )

Options:
A $3.2 \mathrm{~N}$
B $3.2 \times 10^{-16} \mathrm{~N}$
C $1.6 \times 10^{-16} \mathrm{~N}$
D Zero

The correct answer is:
$3.2 \times 10^{-16} \mathrm{~N}$

Solution:
From the question
Velocity of electron $=10^7 \mathrm{~m} / \mathrm{s}$
Magnetic field $B=2 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^2$
Then magnitude of force experienced by the electron i.e.
$$
F=e v \mathrm{~B} \sin \theta
$$
$\{\because V$ and $B$ are prependicular to each other
$$
\begin{aligned}
& =e v \mathrm{~B} \sin 90^{\circ} \\
& =1.6 \times 10^{-19} \times 10^7 \times 2 \times 10^{-4} \times 1 \\
& =3.2 \times 10^{-16} \mathrm{~N}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2013 (Karnataka)
A long straight wire of length $2 \mathrm{~m}$ and mass $250 \mathrm{~g}$ is suspended horizontally in a uniform horizontal magnetic field of $0.7 \mathrm{~T}$. The amount of current flowing through the wire will be $\left(g=9.8 \mathrm{~ms}^{-2}\right)$

Options:
A $2.75 \mathrm{~A}$
B $1.75 \mathrm{~A}$
C $2.45 \mathrm{~A}$
D $2.25 \mathrm{~A}$

The correct answer is:
$1.75 \mathrm{~A}$

Solution:
we know
$\begin{aligned} & F=m g \\ & I \ell B=m g\end{aligned}$
$I=\frac{m g}{\ell B}=\frac{250 \times 10^{-3} \times 9.8}{(2)(0.7)}$
$\begin{aligned} & I=1750 \times 10^{-3} \mathrm{~A} \\ & I=1.75 \mathrm{~A}\end{aligned}$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2023 (Manipur)
A long straight wire of radius a carries a steady current I . the current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B' , at radial distances a2 and 2a respectively, from the axis of the wire is:

Options:
A 14
B 12
C 1
D 4

The correct answer is:
1

Solution:
Inside the wireBy ampereâ²s law
â«Bâ.dlâ=Î¼0ienclosed
â«B.dlcosâ¡0=Î¼0IÏa2.Ïa22
Bâ«dl=Î¼0I4
B2Ïa2=Î¼0I4
B=Î¼0I4Ïa
Outside the wire,
Bâ²=Î¼0I2Ïr=Î¼0I2Ï2a=Î¼0I4Ïa
So, BBâ²=1.

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2016 (Phase 1)
A long straight wire of radius R carries a current i. The magnetic field inside the wire at distance r (r<R), from its centre is expressed as

Options:
A μ0iπR2.r
B 2μ0iπR2.r
C μ0i2πR2.r
D μ0i2πR.r

The correct answer is:
μ0i2πR2.r

Solution:
Using Ampere's law, we have
â®Bâ.dlâ=Î¼0 iin
or BÃ2Ïr=Î¼0iÏR2Ïr2
â´ B=Î¼02Ï.irR2

Topic: Physics, Magnetic Effects of Current, NEET
A long wire bent as shown in the figure carries current I=10 A. If the radius of the semicircular portion is 1 m, the magnetic induction (in Î¼T) at the centre C is

Options:
A π 2 + 4
B Ï 2 + 4
C π 2 - 4
D π 2 - 4

The correct answer is:
π 2 + 4

Solution:
B1 (due to two semi-infinite straight wires) =μ02πIak^=2k^ μT
B2 (semi-circular loop) =μ0I4a-i^=-πi^ μT
B → = B 1 2 + B 2 2

B→=π2+4 μT

Topic: Physics, Magnetic Effects of Current, NEET
A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the center of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of coil will be

Options:
A nB
B n2B
C 3nB
D n3B

The correct answer is:
n2B

Solution:
Suppose the length of wire is l
Then, for the circle of n turns
The magnetic field, B= Î¼ 0 In 2R , where R is the radius of the circle
Then Bs=Î¼0 I2R =B......(i)
Here, 2ÏR=l .......(ii)
Now, if the same wire is turned into a coil of n turns of radius a then,
2ÏaÃn=l .......(iii)
Thus, 2ÏR=2ÏaÃn
â a=Rn
Now, the magnetic field for the coil of n turns, is
Bn=Î¼0I2aÃn
=Î¼0I2RÃn2 âµ a=Rn
=n2B

Topic: Physics, Magnetic Effects of Current, NEET
A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be

Options:
A nB
B n2B
C 2nB
D 2n2B

The correct answer is:
n2B

Solution:
B=μ0I2r.

When n turns are made, radius becomes r′.

n×2πr′=2πr⇒r′=rn.

Now, B′=μ0nI2r′=n2μ0I2r=n2B.

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2016 (Phase 2)
A long wire is divided into 2n parts. Then n parts are connected in series and the other n parts are connected in parallel. Both the combinations are given the same voltage supply. The ratio of heat produced in series to the parallel combination will be

Options:
A 1:1
B 1:n2
C 1:n4
D n2:1

The correct answer is:
1:n2

Solution:
The resistance of each part is r=R/2n
Rseries =R2n× n =R2
The heat developed in series combination
=V2R/2 J s-1
Rparallel=rn= R/2nn= R2n2
The heat produced in parallel combination =V2R/2n2 J s-1
∴ HseriesHparallel= 2V2R× R2n2V2= 1:n2

Topic: Physics, Current Electricity, NEET
A loop ABCD has current I=10 A, as shown in the figure. AD and BC are circular arcs with centre at O, for both. The magnetic field at point O is

Options:
A 10-4Â T
B 10-5 T
C 1.5Ã10-5Â T
D 2Ã10-5Â T

The correct answer is:
10-5 T

Solution:
Magnetic field due to AB and CD are zero at O
Magnetic field due to BC
BBC=Î¼04Ï103Ã10-2ÃÏ4 (outwards)
Simplifying field due to DA
BSP=Î¼04Ï105Ã10-2ÃÏ4 (inwards)
Net field B=Î¼o4Ï10Ï4Ã10-213-15
=1.0Ã10-5 T

Topic: Physics, Magnetic Effects of Current, NEET
A loop ABCD is moving with velocity v towards the right. The magnetic field is 2 T. Loop is connected to a resistance of 9 Î©. If the steady current of 2 A flows in the loop, then the value of v, if loop has resistance of 4 Î©, is (Given AD = 30 cm)

Options:
A 86.7 m s-1

B 30 m s-1

C 33.33 m s-1

D 20 m s-1

The correct answer is:
86.7 m s-1

Solution:
When a straight conductor moves with velocity v perpendicular to the magnetic field then emf induced is given by V=Bvl. Here, it will be V=Bv(l sin θ).
Potential difference is V=Bv (l sin θ)
V=2×v×30100×12=0·3v
Now, Current=VR
⇒ 2=0·3V13
⇒ v=86.7 m s-1

Topic: Physics, Electromagnetic Induction, NEET
A loop of capillary running parallel to the Henle's loop is known as ___.

Options:
A Vasa recta
B Vasa efferentia
C Vasa deferentia
D Henle's loop

The correct answer is:
Vasa recta

Solution:
A loop of capillary running parallel to the Henle's loop is known as Vasa recta

Topic: Biology, Excretory Products and their Elimination, NEET
A machine gun fires a bullet of mass 40 g with a velocity 1200 m sâ1. The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most?

Options:
A One
B Four
C Three
D Two

The correct answer is:
Three

Solution:
According to Newton's second law
F=âpât
F=nmvt
F= mnv
144= 40 Ã 10-3 Ã n Ã 1200
n=1444Ã12=3

Topic: Physics, Center of Mass Momentum and Collision, NEET
A machine which is 75% efficient uses 12 J of energy in lifting up a 1 kg mass through a certain distance. The mass is then allowed to fall through, that distance. The velocity of the ball at the end of its fall is

Options:
A 24 m s-1
B 32 m s-1
C 18 m s-1
D 3 m s-1

The correct answer is:
18 m s-1

Solution:
Useful work =75100Ã12 J=9J
Now, 12Ã1Ãv2=9 or v=18 m s-1

Topic: Physics, Work Power Energy, NEET
A magnet is suspended horizontally in the earth's magnetic field. When it is displaced and released, it oscillates in a horizontal plane with a period T. If a piece of wood of same moment of inertia as the magnet is attached to the magnet, the new period of oscillation of the system would be

Options:
A T3
B T2
C T2
D 2T

The correct answer is:
2T

Solution:
T=2ÏlMBH
T′=2Ï2lMBH=2T

Topic: Physics, Magnetic Properties of Matter, NEET
A magnet is suspended in such a way that it oscillates in the horizontal plane. It makes 20 oscillations per minute at a place where dip angle is 30o and 15 oscillations per minute at a place where dip angle is 60o . Ratio of the total Earth's magnetic field at the two places is

Options:
A 33:8
B 16:93
C 4 :9
D 23:9

The correct answer is:
16:93

Solution:
If θ is dip angle, then time period of oscillation is
T=2πIMBcosθ
∴ T2T1=B1cos⁡θ1B2cos⁡θ2
∴ B1B2=T22cos⁡θ2T12cos⁡θ1
B1B2=11521202.cos⁡60ocos⁡30o
B1B2=1693

Topic: Physics, Magnetic Properties of Matter, NEET
A magnet is suspended lengthwise from a spring and while it oscillates, the magnet moves in and out of the coil C connected to a galvanometer G. Then, as the magnet oscillates

Options:
A G shows no deflection
B G shows deflection on one side
C deflection of G to the left and right has constant amplitude
D deflection of G to the left and right has decreasing amplitude

The correct answer is:
deflection of G to the left and right has decreasing amplitude

Solution:
As a given pole ( N or S ) of suspended magnet goes into the coil and comes out of it, current is induced in the coil in two opposite directions. Therefore, galvanometer deflection goes to left and right both. As the amplitude of oscillation of magnet goes on decreasing, so does the amplitude of deflection.

Topic: Physics, Electromagnetic Induction, NEET
A magnet of a magnetic moment M is in the form of a quadrant of a circle. If it is straightened, its new magnetic moment will be

Options:
A Mπ2
B M2
C 2Mπ
D Mπ22

The correct answer is:
Mπ22

Solution:
Magnetic length is minimum length between north pole and south pole
M′=Mθ2sin⁡θ2
Here θ=π2
M' = Mπ22

Topic: Physics, Magnetic Properties of Matter, NEET
A magnet of dipole moment 2 A m2 is deflected through 30Â° from magnetic meridian. The required deflecting torque is BH=0.4Ã10-4 T

Options:
A 0.4×10-4 N m
B 0.4 N m
C 0.2×10-4 N m
D 0.8 N m

The correct answer is:
0.4×10-4 N m

Solution:
When a magnet is placed in an external magnetic field it experiences torque
τ=MBsin⁡θ
=2×0.4×10-4×sin⁡30°
=0.4×10-4 N m

Topic: Physics, Magnetic Properties of Matter, NEET
A magnet of moment 4 A m2 is kept suspended in a magnetic field of induction 5Ã10-5 T. The work done in rotating it through 1800 is

Options:
A 4Ã10-4 J
B 5Ã10-4Â J
C 2Ã10-4Â J
D 10-4Â J

The correct answer is:
4Ã10-4 J

Solution:
W=−M.B(cos θ f −cos θ i )
=( − )4×5× 10 −5 (cos180−cos0)
=−20× 10 −5 ( −1−1 )
=4×10−4 J

Topic: Physics, Magnetic Properties of Matter, NEET
A magnetic dipole is placed at right angles to the direction of magnetic induction B. If it is rotated through an angle of 180°, then the work done is

Options:
A MB
B 2 MB
C -2 MB
D zero

The correct answer is:
zero

Solution:
Work done in rotating a dipole
W=MB cos⁡θ1-cos⁡θ2
To rotate it from 90° to 0°
θ1=90° θ2=0°
W=MBcos⁡90°-cos⁡0°
W1=MB0-1=-MB
W1=-MB .........(i)
To rotate it from 0° to -90° in same manner.
θ1=0° θ2=-90°
W2=MBcos⁡0°-cos⁡-90°
W2=MB ........(ii)
Total workdone
W=W1+W2
W=-MB+MB
W=0

Topic: Physics, Magnetic Properties of Matter, NEET
A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60Â° and one of the fields has a magnitude of 1.2Ã10-2 T. If the dipole comes to stable equilibrium at an angle of 15Â° with this field, what is the magnitude of the other field?

Options:
A 4.4×10-3

B 4.4Ã10-4

C 4.8Ã10-3
D 4.4Ã103

The correct answer is:
4.4×10-3

Solution:
Let one of the magnetic fields is B1 and other is B2. Angle between B1 and B2 is 60o.
Given, B1 = 1.2 x 10-2 T
Dipole is in equilibrium at an angle 15o from B1 or 60o - 15o = 45o from B2
Torque on dipole due to magnetic field B1
Ï 1 = M Ã B 1 sin 1 5 o ......(i)

where, M is the magnetic moment. Torque on dipole due to magnetic field B2
Ï 2 = M Ã B 2 sin 4 5 o .....(ii)
As dipole is in the equilibrium, then
Ï 1 = Ï 2
M X B1 sin 15o = M x B2 sin 45o [From Eqs. (i) and (ii)]
1 . 2 Ã 1 0 - 2 Ã sin 1 5 o sin 4 5 o = B 2
B 2 = 1 . 2 Ã 1 0 - 2 Ã 0 . 2 5 8 8 0 . 7 0 7 1 = 4 . 4 Ã 1 0 - 3 T
Thus, the magnitude of the other field is 4.4. x 10-3 T.

Topic: Physics, Magnetic Effects of Current, NEET
A magnetic dipole of magnetic moment 6Ã10-2A m2 and moment of inertia 12Ã10-6 kg m2 performs oscillations in a magnetic field of 2Ã10-2T. The time taken by the dipole to complete 20 oscillations is Ïâ3

Options:
A 18 s
B Â 6s
C 36 s
D 12 s

The correct answer is:
12 s

Solution:
Given,
Magnetic moment M=6Ã10-2A m2
Moment of inertia I=12Ã10-6 kg m2
Magnetic field B=2Ã10-2 T
We know that,
Time T=2ÏIMB
=2Ï12Ã10-66.Ã10-2Ã2Ã10-2
=2Ï12Ã10-212
=2ÏÃ10-1 s
For 20 oscillations of tâ12 s

Topic: Physics, Magnetic Effects of Current, NEET
A magnetic field B is confined to a region r a ≤ and points out of the paper (the z-axis), r = 0 being the centre of the circular region. A charged ring (charge = Q) of radius b, b > a and mass m lies in the x-y plane with its centre at the origin. The ring is free to rotate and is at rest. The magnetic field is brought to zero in time ∆t. Find the angular velocity ω of the ring after the field vanishes.

Options:
A $\omega =\frac{QB{{a}^{2}}}{m{{b}^{2}}}$
B $\omega =\frac{QB{{a}^{2}}}{2m{{b}^{2}}}$
C $\omega =\frac{QB{{a}^{2}}}{3m{{b}^{2}}}$
D $\omega =\frac{QB{{a}^{2}}}{4m{{b}^{2}}}$

The correct answer is:
$\omega =\frac{QB{{a}^{2}}}{2m{{b}^{2}}}$

Solution:
Correct Option is : (B)
$\omega =\frac{QB{{a}^{2}}}{2m{{b}^{2}}}$

Topic: Chemistry, Electromagnetic Induction, NEET
A magnetic needle free to rotate in a horizontal plane is placed at the centre of a circular current-carrying coil whose axis is perpendicular to the magnetic meridian at that place. It is also known that the magnetic declination at this place is zero and in this condition, the magnetic needle is pointing towards the north-west. Now, if we reverse the direction of current in the coil, then the magnetic needle will

Options:
A point north-west
B point north-east
C point south-east
D point south-west

The correct answer is:
point north-east

Solution:
If there is no current in the coil, then the needle would point north but due to the magnetic field of the coil, it is pointing north-west. This means the magnetic field due to the coil is towards west and it is equal to the horizontal component of earth's magnetic field. So when we reverse the direction of current the magnetic field due to the coil will be towards east and the magnetic needle will point north-east.

Topic: Physics, Magnetic Properties of Matter, NEET
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22 0 with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth's magnetic field at the place. cos22Â°=0.9272

Options:
A 0.38 G
B 0.35 G
C 0.30 G
D 0.40 G

The correct answer is:
0.38 G

Solution:
Given, angle of dip Î´ = 2 2 o
Horizontal component of the earth's magnetic field H = 0.35 G
Let the magnitude of the earth's magnetic field at the place is R .
Using the formula, H = R cos Î´
or R = H cos Î´ = 0 . 3 5 cos 2 2 o = 0 . 3 5 0 . 9 2 7 2 = 0 . 3 8 G
Thus, the value of the earth's magnetic field at that place is 0.38 G .

Topic: Physics, Magnetic Properties of Matter, NEET
A magnetic needle is kept in a non-uniform magnetic field. It experiences

Options:
A a force and a torque
B a force but not a torque
C a torque but not a force
D neither a force nor a torque

The correct answer is:
a force and a torque

Solution:
Force as well as torque both acts on a magnetic needle kept in a non-uniform magnetic field.

Topic: Physics, Magnetic Effects of Current, NEET
A magnetic needle is kept in a non-uniform magnetic field. It experiences

Options:
A a force and a torque
B a force but not a torque
C a torque but not a force
D neither a force nor a torque

The correct answer is:
a force and a torque

Solution:
As the field is non uniform therefore force experienced by both the poles has different magnitude, so there will be a net force and a torque.

Topic: Physics, Magnetic Properties of Matter, NEET
A magnetic wire of dipole moment 4Ï Am2 is bent in the form of semicircle. The new magnetic moment is

Options:
A 4π A m2
B 8π A m2
C 4 A m2
D None of these

The correct answer is:
4 A m2

Solution:
Here, Mi=4π
mL=4π
m=4πL
The new distance between poles is
L1=2r=2L2π=Lπ
∴ Mf=mL1=mLπ=4ππ=4 A m2

Topic: Physics, Magnetic Properties of Matter, NEET