Question:
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 isΒ
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 isΒ
Updated On: Mar 19, 2025
- zero in the region r < a and inversely proportional to r in the region r > a
- inversely proportional to r in the region r < a and uniform throughout in the region r > a
- directly proportional to r in the region r < a and inversely proportional to r in the region r > aΒ
- uniform in the region r < a and inversely proportional to distance r from the axis, in the region r > a
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
It is a case of solid infinite current carrying wire.
Using ampere circuital law,
CASE I: if r β€ R π΅ = π_0π/2ππ ^2 π
CASE II: r>R π΅ = π0π /2πr
Using ampere circuital law,
CASE I: if r β€ R π΅ = π_0π/2ππ ^2 π
CASE II: r>R π΅ = π0π /2πr
Was this answer helpful?
0
0
Top Questions on Moving charges and magnetism
- (a) State Faradayβs law of electromagnetic induction and explain the role of the negative sign in its expression.
(b) Explain, with an example, that Lenzβs law is consistent with the law of conservation of energy.- CBSE CLASS XII - 2025
- Physics
- Moving charges and magnetism
- (i) A rectangular loop of sides \( l \) and \( b \) carries a current \( I \) clockwise. Write the magnetic moment \( \vec{m} \) of the loop and show its direction in a diagram.
(ii) The loop is placed in a uniform magnetic field \( \vec{B} \) and is free to rotate about an axis which is perpendicular to \( \vec{B} \). Prove that the loop experiences no net force, but a torque \( \vec{\tau} = \vec{m} \times \vec{B} \).- CBSE CLASS XII - 2025
- Physics
- Moving charges and magnetism
- Assertion (A): The deflection in a galvanometer is directly proportional to the current passing through it.
Reason (R): The coil of a galvanometer is suspended in a uniform radial magnetic field.- CBSE CLASS XII - 2025
- Physics
- Moving charges and magnetism
- A particle of charge \( q \) is moving with a velocity \( \vec{v} \) at a distance \( d \) from a long straight wire carrying a current \( I \) as shown in the figure. At this instant, it is subjected to a uniform electric field \( \vec{E} \) such that the particle keeps moving undeviated. In terms of unit vectors \( \hat{i}, \hat{j}, \) and \( \hat{k} \), find:
the magnetic field \( \vec{B} \),
the magnetic force \( \vec{F}_m \), and
the electric field \( \vec{E} \) acting on the charge.
- CBSE CLASS XII - 2025
- Physics
- Moving charges and magnetism
- A charged particle is moving in a circular path with velocity \( \vec{V} \) in a uniform magnetic field \( \vec{B} \). It is made to pass through a sheet of lead and as a consequence, it loses one-half of its kinetic energy without change in its direction. How will:
1. the radius of its path change? 2. its time period of revolution change?- CBSE CLASS XII - 2025
- Physics
- Moving charges and magnetism
View More Questions
Questions Asked in JEE Main exam
- A force of 49 N acts tangentially at the highest point of a sphere (solid of mass 20 kg) kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is:
- JEE Main - 2025
- Quantum Mechanics
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
- JEE Main - 2025
- Relations and functions
- If \( \alpha + i\beta \) and \( \gamma + i\delta \) are the roots of the equation \( x^2 - (3-2i)x - (2i-2) = 0 \), \( i = \sqrt{-1} \), then \( \alpha\gamma + \beta\delta \) is equal to:
- JEE Main - 2025
- Complex numbers
- The sum of all local minimum values of the function \( f(x) \) as defined below is:
\[ f(x) = \begin{cases} 1 - 2x & \text{if } x < -1, \\[10pt] \frac{1}{3}(7 + 2|x|) & \text{if } -1 \leq x \leq 2, \\[10pt] \frac{11}{18}(x-4)(x-5) & \text{if } x > 2. \end{cases} \]- JEE Main - 2025
- Functions
- Let \( \langle a_n \rangle \) be a sequence such that \( a_0 = 0 \), \( a_1 = \frac{1}{2} \), and \( 2a_{n+2} = 5a_{n+1} - 3a_n \).n= 0,1,2,3.... Then \( \sum_{k=1}^{100} a_k \) is equal to:
- JEE Main - 2025
- Sequences and Series
View More Questions