The figure given below shows a long straight solid wire of circular cross-section of radius 'a' carrying steady current I. The current I is uniformly distributed across its cross-section. The plot whi

For a long straight solid wire carrying a steady current that is uniformly distributed across its cross-section: 1. Inside the wire (r a), the magnetic field is inversely proportional to the distance from the axis (B propto (1/r)). This represents a rectangular hyperbola. The correct graph must show a linear increase starting from the origin up to r=a, followed by a smooth 1/r hyperbolic curve decaying towards zero for r>a. The plot matching this criteria is option 1. Solid cylinder uniform curr

The figure given below shows a long straight solid wire of circular cross-section of radius 'a' carrying steady current I. The current I is uniformly distributed across its cross-section. The plot whi

For a long straight solid wire carrying a steady current that is uniformly distributed across its cross-section: 1. Inside the wire (r a), the magnetic field is inversely proportional to the distance from the axis (B propto (1/r)). This represents a rectangular hyperbola. The correct graph must show a linear increase starting from the origin up to r=a, followed by a smooth 1/r hyperbolic curve decaying towards zero for r>a. The plot matching this criteria is option 1. Solid cylinder uniform curr

The figure given below shows a long straight solid wire of circular cross-section of radius 'a' carrying steady current…

Year	2024	Total
Questions	3	3

Year

2024

Total

Questions

Solution & Explanation

Related Formula

Binside = (μ₀ I r)/(2π a²) (r < a)

Boutside = (μ₀ I)/(2π r) (r > a)

Core Logic

For a long straight solid wire carrying a steady current that is uniformly distributed across its cross-section:

Inside the wire ( $r < a$ $r < a$), the magnetic field is directly proportional to the distance from the axis ( $B \propto r$ $B \propto r$). This represents a straight line passing through the origin.
Outside the wire ( $r > a$ $r > a$), the magnetic field is inversely proportional to the distance from the axis ( $B \propto (1)/(r)$ $B \propto \frac{1}{r}$). This represents a rectangular hyperbola.

Step 1: Graph Identification

The correct graph must show a linear increase starting from the origin up to $r=a$ $r=a$, followed by a smooth $1/r$ $1/r$ hyperbolic curve decaying towards zero for $r>a$ $r>a$. The plot matching this criteria is option 1.

Pattern Recognition

Solid cylinder uniform current: Inside $B \propto r$ $B \propto r$ (linear), outside $B \propto 1/r$ $B \propto 1/r$ (curve). Hollow cylinder: Inside $B = 0$ $B = 0$, outside $B \propto 1/r$ $B \propto 1/r$.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

More Moving Charges and Magnetism Previous-Year Questions

Q1 neet_2026_03_may_morning Magnetic Field on the Axis of a Circular Current Loop

100

-turn closely wound circular coil of radius

5 ~cm

has a magnetic field

3.14 \times 10⁻³ ~T

at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of the coil are, respectively - (Take

μ₀ = 4π \times 10⁻⁷ ~Tm/A

)

A. $2.5 ~A, 2 ~Am²$ $2.5 \mathrm{~A}, 2 \mathrm{~Am}^2$
B. $2.5 ~A, 20 ~Am²$ $2.5 \mathrm{~A}, 20 \mathrm{~Am}^2$
C. $2 ~A, 4 ~Am²$ $2 \mathrm{~A}, 4 \mathrm{~Am}^2$
D. $2 ~A, 10 ~Am²$ $2 \mathrm{~A}, 10 \mathrm{~Am}^2$

Solution

Related Formula

B₀ = (μ₀ N i)/(2 R)

M = N i A = N i π R²

Core Logic

First, calculate the current $i$ $i$ using the magnetic field formula for a circular loop:

i = (2 R B₀)/(μ₀ N)

Substitute the given values: $R = 5 \times 10⁻² ~m$ $R = 5 \times 10^{-2} \mathrm{~m}$ $B₀ = 3.14 \times 10⁻³ ~T$ $B_0 = 3.14 \times 10^{-3} \mathrm{~T}$ $N = 100$ $N = 100$

i = 2 \times 5 \times 10⁻² \times 3.14 \times 10⁻³4π \times 10⁻⁷ \times 100

Using $π \approx 3.14$ $\pi \approx 3.14$, we get:

i = 10 \times 10⁻² \times π \times 10⁻³4π \times 10⁻⁵ = 10⁻⁴4 \times 10⁻⁵ = 2.5 ~A

Step 1: Calculate Magnetic Moment

Now, compute the magnetic moment $M$ $M$:

$M = N i π R²$ $M = N i \pi R^2$

M = 100 \times 2.5 \times 3.14 \times (5 \times 10⁻²)²

M = 250 \times 3.14 \times 25 \times 10⁻⁴

M = 6250 \times 3.14 \times 10⁻⁴ \approx 2 ~Am²

Pattern Recognition

Recognize that $π \approx 3.14$ $\pi \approx 3.14$ cancels out nicely when substituted directly against the given magnetic field value.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q12 neet_2026_03_may_morning Conversion of Galvanometer to Ammeter

A galvanometer of resistance

100Ω

gives full scale deflection for a current of

1 ~mA

. It is converted into an ammeter of range

0--10 ~A

. The shunt required is:

A. $0.01Ω$ $0.01\Omega$
B. $0.10Ω$ $0.10\Omega$
C. $0.001Ω$ $0.001\Omega$
D. $1.0Ω$ $1.0\Omega$

Solution

Related Formula

IS RS = IG RG

RS = (IG RG)/(I - IG)

Core Logic

Given: Galvanometer resistance $RG = 100 Ω$ $R_G = 100 \Omega$ Full scale deflection current $IG = 1 ~mA = 0.001 ~A$ Target range current $I = 10 ~A$ $I = 10 \mathrm{~A}$

The shunt resistance $RS$ $R_S$ is connected in parallel with the galvanometer. Since they are in parallel, the potential difference across them is equal.

$VS = VG$ $V_S = V_G$

(I - IG)RS = IG RG

Because $IG (0.001 ~A)$ $I_G (0.001 \mathrm{~A})$ is very small compared to $I (10 ~A)$ $I (10 \mathrm{~A})$, we can approximate $(I - IG) \approx I = 10 ~A$ $(I - I_G) \approx I = 10 \mathrm{~A}$.

Step 1: Calculate Shunt Resistance

10 \times RS = 0.001 \times 100

10 \times RS = 0.1

RS = (0.1)/(10) = 0.01 Ω

Pattern Recognition

Ammeter conversion relies heavily on the fact that the vast majority of current passes through the very low resistance shunt parallel path.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Solution & Explanation

Related Formula

Core Logic

Step 1: Graph Identification

Pattern Recognition

Chapter Mix

Reference Study Guides

More Moving Charges and Magnetism Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Calculate Magnetic Moment

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Shunt Resistance

Pattern Recognition

Chapter Mix

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Solution & Explanation

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Step 1: Graph Identification

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Reference Study Guides

More Moving Charges and Magnetism Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Calculate Magnetic Moment

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Shunt Resistance

Pattern Recognition

Chapter Mix

More Moving Charges and Magnetism Questions — neet_2026_03_may_morning

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