29 NEET previous-year questions on Moving Charges and Magnetism, each with the correct answer and a step-by-step solution. Sourced directly from official NEET papers across every booklet code.
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499 1 G
500 499 G
500 1 G
499 500 G
Solution
In an ammeter, the shunt resistance is given by , where is the ratio of the total current to the current through the galvanometer. Here, . Substituting, . The total resistance of the ammeter is , so option (c) is correct.
π 2 1 0 I I d 2 μ
d 2 μ 0 π (I 1 + I 2 )
) I I ( d 2 μ 2 2 2 1 0 − π
2 / 1 2 2 2 1 0 ) I I ( d 2 μ + π
Solution
The magnetic field at point P due to each wire can be calculated using the Biot-Savart law. For two perpendicular wires carrying currents and , the fields add vectorially. The magnitude of the resultant magnetic field is given by . Therefore, the correct option is (d).
abcd
adcb
The current will reverse its direction as the electron goes past the coil
No current induced
Solution
When e− comes closer the induced current will be anticlockwise When e− comes farther induced current will be clockwise.
4 MeV
0.5 MeV
1..5 MeV
1 MeV
Solution
R= mV qB =√2m(kE) qB Since R is same so KE ∝ q2 m So KE of α particle will be (2)2 4 =same=1 MeV
2μ0Ii 3π
μ0Ii 2π
2μ0IiL 3π
μ0IiL 2π
Solution
F1 = μ0IiL 2πL 2 = μ0Ii π 2 = μ0IiL 2π3L 2 = μ0Ii 3π ∴ Fnet = F1 − F2 Fnet = 2 3 μ0Ii π
1 4
1 2
1
4
Solution
Inside the wire By ampere′s law ∫B⃗⃗ .dl⃗⃗⃗ = μ0(ienclosed) ∫B. dlcos0 = μ0 ( I πa2 .π(a 2) 2 ) B∫ dl= μ0 I 4 B(2π(a 2))= μ0I 4 www.vedantu.com 18 B = μ0I 4πa Outside the wire, B′ = μ0I 2πr = μ0I 2π(2a)= μ0I 4πa So, B B′ = 1.
4 H
3 H
2 H
1 H
Solution
ϕ =Li 1000× 4 ×10−3 = L4 1 = L
9.1 μJ
4.55 μJ
2.3 μJ
1.15 μJ
Solution
W = MB (cosθ 1 – cosθ2) www.vedantu.com 2 When it is rotated by angle 180º then W = 2MB W = 2 (NIA)B = 2 × 250 × 85 × 10 –6[1.25 × 2.1 × 10 –4] × 85 × 10–2 = 9.1 μJ
2 0 2 I d μ π
2 02 I d μ π
2 02 I d μ π
2 0 2 I d μ π
Solution
Force between BC and AB will be same in magnitude. 90° C B A F d F d 2 0 2BC BA IFF d μ== π 2 BCFF= 2 02 2 I d μ= π 2 0 2 IF d μ= π www.vedantu.com 8
32 π?μC
16 μC
32 μC
16 π?μC
Solution
dN dt φε=- Nd RR d t εφ = Ndq d R=φ ()NQ R ΔφΔ= totalQ R ΔφΔ= ()NBA R= 2 0ni r R μπ= Putting values 72 2 4 10 100 4 (0.01 ) 10 -π× × × ×π×= π 32 CQΔ= μ
250 Ω
25 Ω
40 Ω
500 Ω ACHLA/AA/Page 6 SPACE FOR ROUGH WORK English
Solution
The voltage sensitivity is given by , where is the current sensitivity and is the resistance of the galvanometer. Substituting the given values, . However, the correct calculation should be , so option (a) is correct.
1 J
3 J
30 kJ
2 J
Solution
Work = 3 4×100× 20 100× 20 100 = 3J
zero
𝐴ω 2
𝐴ω
𝐴ω2 2
Solution
Average velocity = 𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 = 0
6.28 × 10 − 4 T
3.14 × 10 − 4 T
6.28 × 10 − 5 T
3.14 × 10 − 5 T
Solution
The magnetic field inside a solenoid is given by . Substituting the values, , so option (a) is correct.
4 × 10 − 20 N
8 π× 10 − 20 N
4 π× 10 − 20 N
8 × 10 − 20 N
Solution
The magnetic field due to an infinitely long straight conductor is . Substituting, . The force on the electron is , so option (c) is correct.
Solution
For a thick current-carrying cable, the magnetic field inside the cable () is proportional to , and outside the cable () it is inversely proportional to . This is represented by a linear increase inside the cable and a hyperbolic decrease outside, as shown in option (a).
8 8 6 i j k ∧ ∧ ∧ − − −
6 6 8 i j k ∧ ∧ ∧ − − −
8 8 6 i j k ∧ ∧ ∧ + −
6 6 8 i j k ∧ ∧ ∧ + −
Solution
Given , with , , and . Using the cross product, . Equating components, , , and . Solving these, , , and , so . Option (d) is correct.
6.28 × 10–2 T
12.56 × 10–2 T
12.56 × 10–4 T
6.28 × 10–4 T
Solution
We know, magnetic field at centre of solenoid 00= µ= µλ NB l nl = λ Nn = 4π × 10–7 × 100 × 103 × 1 3 100 10− = n = 4π × 10–2 T B = 12.56 × 10–2 T
Both Statement I and Statement II are correct
Both Statement I and Statement II are incorrect
Statement I is correct and Statement II is incorrect
Statement I is incorrect and Statement II is correct
Solution
According to Biot -Savart’s law 0 34 µ ×= π ρ ρρ Idl rdB r which is applicable for infinitesimal element. It is analogous to Coulomb’s law, where Idl ρ is vector source and electric field is produced by scalar source q. Here statement I is correct and statement II is incorrect.
Uniform and remains constant for both the regions.
A linearly increasing function of distance upto the boundary of the wire and then linearly decreasing for the outside region.
A linearly increasing function of distance r upto the boundary of the wire and then decreasing one with 1 r dependence for the outside region.
A linearly decreasing function of distance upto the boundary of the wire and then a linearly increasing one for the outside region.
Solution
For solid wire Inside point 2 0 2 2 IrB Rr µ= ×π 0 2 2 Ir R µ= ×π Br∝ Outside point 0 2 IB r µ= π 1B r∝
0.25 A
1.5 A
1 A
2 A
Solution
φB = NBA cosωt ( sin )Bd NBA tdt −φε= =− ω− ω sinNBA tε= ω ω max max NBAi RR ε ω= = 521000 2 10 (10) 2 12.56 −× × ×π ×= = 1 A
5π2
128π2
50π2
1280π2
Solution
Time period of Oscillation, 2 IT MB=π ⇒ 61 9.8 1024 0.049 M −×=π × ⇒ 62 3 1 9.8 10416 49 10M − − ×= π × ×× ⇒ 26 3 4 9.8 10 16 49 10 M − − π × ×=× × = 234 9.8 16 10 49 −π × × × = 12.8π2 × 10–3 × 10–2 × 102 = 1280π2 × 10–5 Am2
44 mT
4.4 T
4.4 mT
44 T
Solution
The magnitude of magnetic field due to circular coil of N turns is given by 0 2 μ=C iNB R 74 10 7 100 2 0.1 −π× × ×= × = 4.4 × 10–3 T = 4.4 mT
A-II, B-III, C-IV, D-I
A-II, B-I, C-III, D-IV
A-III, B-II, C-I, D-IV
A-IV, B-III, C-II, D-I
Solution
(Material) (Susceptibility (χ)) Diamagnetic (II) 0 > χ ≥ –1 Ferromagnetic (III) χ >> 1 Paramagnetic (IV) 0 < χ < ε Non-magnetic (I) χ = 0 - 16 - NEET (UG)-2024 (Code-Q1)
M
2 M
2M
3 M
Solution
M = ml. 2 sin302 llΔ = ° = 2 l M′ = ml/2 = M/2
B and D only
A and C only
A, C and D only
C only
Solution
A. A magnetic pole will repel or attract magnetic sheet so force is need. B. If sheet is non-magnetic, no force needed. C. If it is conducting, then there will be addy current in sheet, which opposes the motion. So forces is needed move sheet with uniform speed. D. The non-conducting and non-polar sheet do not interact with magnetic field of magnet. CHEMISTRY SECTION-A
is perpendicular to and its magnitude is V m
is perpendicular to and its magnitude is V m
is parallel to and its magnitude is V m
is parallel to and its magnitude is V m
Solution
For no deflection of electron,
V m
1 : 4
1 : 2
2 : 1
4 : 1
Solution
Magnetic moment of current carrying circular loop =
Solution
Magnetic force =
for
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