(C) For a long straight wire with uniform current distribution:$1$. Inside the wire $(r < a)$: Using Ampere's circuital law,$\oint B \cdot dl = \mu_0

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Directly proportional to $r$ in the region $r a$

(C) For a long straight wire with uniform current distribution:$1$. Inside the wire $(r $2$. Outside the wire $(r > a)$: Using Ampere's circuital law,$B(2\pi r) = \mu_0 I$,which gives $B = \frac{\mu_0 I}{2\pi r}$. Therefore,$B \propto \frac{1}{r}$.

Class 12 Physics Moving Charges and Magnetism Ampere’s circuital law and its application (Solenoid and Toroid)

Medium

$A$ long straight wire of circular cross-section (radius $a$) is carrying steady current $I$. The current $I$ is uniformly distributed across this cross-section. The magnetic field is

JEE MAIN 2023 All JEE MAIN

A
Zero in the region $r < a$ and inversely proportional to $r$ in the region $r > a$
B
Inversely proportional to $r$ in the region $r < a$ and uniform throughout in the region $r > a$
C
Directly proportional to $r$ in the region $r < a$ and inversely proportional to $r$ in the region $r > a$
D
Uniform in the region $r < a$ and inversely proportional to distance $r$ from the axis,in the region $r > a$

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Class 12

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Moving Charges and Magnetism

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Ampere’s circuital law and its application (Solenoid and Toroid)

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The magnetic energy stored in a long solenoid of area of cross-section $A$ in a small region of length $L$ is

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The number of turns per unit length of a long solenoid is $10 \, \text{turns/cm}$. If its average radius is $5 \, \text{cm}$ and it carries a current of $10 \, \text{A}$, then the ratio of flux densities obtained at the centre and at the end on the axis will be:

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Inductance per unit length near the middle of a long solenoid is ($\mu_0=$ permeability of free space,$n=$ number of turns per unit length,$d=$ the diameter of the solenoid).

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$A$ solenoid of length $50 \,cm$ having $100$ turns carries a current of $2.5 \,A$. The magnetic field at one end of the solenoid is

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Using Ampere's circuital law,derive the expression for the magnetic field at a perpendicular distance $r$ from an infinitely long current-carrying wire. Explain it.

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$A$ long straight wire of circular cross-section (radius $a$) is carrying steady current $I$. The current $I$ is uniformly distributed across this cross-section. The magnetic field is

Similar Questions

The magnetic energy stored in a long solenoid of area of cross-section $A$ in a small region of length $L$ is

The number of turns per unit length of a long solenoid is $10 \, \text{turns/cm}$. If its average radius is $5 \, \text{cm}$ and it carries a current of $10 \, \text{A}$, then the ratio of flux densities obtained at the centre and at the end on the axis will be:

Inductance per unit length near the middle of a long solenoid is ($\mu_0=$ permeability of free space,$n=$ number of turns per unit length,$d=$ the diameter of the solenoid).

$A$ solenoid of length $50 \,cm$ having $100$ turns carries a current of $2.5 \,A$. The magnetic field at one end of the solenoid is

Using Ampere's circuital law,derive the expression for the magnetic field at a perpendicular distance $r$ from an infinitely long current-carrying wire. Explain it.

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