A long solenoid of radius 1,mm has 100 turns | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The magnetic field $B$ at the centre of a long solenoid is given by the formula $B = \mu_0 n i$,where $n$ is the number of turns per unit length a

Vedclass · Accepted Answer

$12.56 	imes 10^{-2}\,T$

Vedclass · Answer

(A) The magnetic field $B$ at the centre of a long solenoid is given by the formula $B = \mu_0 n i$,where $n$ is the number of turns per unit length and $i$ is the current.Given:Number of turns per unit length $n = 100 	ext{ turns/mm} = 100 	imes 10^3 	ext{ turns/m} = 10^5 	ext{ turns/m}$.Current $i = 1\,A$.Permeability of free space $\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T}\,	ext{m/A}$.Substituting the values:$B = (4\pi 	imes 10^{-7}) 	imes (10^5) 	imes (1)$$B = 4\pi 	imes 10^{-2} 	ext{ T}$$B \approx 4 	imes 3.14 	imes 10^{-2} 	ext{ T} = 12.56 	imes 10^{-2} 	ext{ T}$.

$A$ 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:

Similar Questions

Ampere's circuital law is associated with which law?

$A$ toroid has an iron core with an internal magnetic field of $10 \pi \text{ mT}$,when the current in the winding of $1500 \text{ turns/m}$ is $10 \text{ A}$. Determine the field due to magnetization $(\mu_0 = 4 \pi \times 10^{-7} \text{ H m}^{-1})$.

The magnetic induction along the axis of a toroidal solenoid is independent of

Write the equation of magnetic field on the axis of a current-carrying finite solenoid.

Assertion : If the current in a solenoid is reversed in direction while keeping the same magnitude,the magnetic field energy stored in the solenoid decreases.
Reason : Magnetic field energy density is proportional to the square of the current.

Vedclass Test Series

Exam Paper Generator

Online Exam Module