Class 12PhysicsMoving Charges and MagnetismAmpere’s circuital law and its application (Solenoid and Toroid)
Medium$A$ solenoid of $1000$ turns is wound uniformly on a glass tube $4 \text{ m}$ long and $0.3 \text{ m}$ in diameter. The magnetic intensity at the centre of the solenoid when a current of $4 \text{ A}$ flows through it is
- A$2 \times 10^{3} \text{ A/m}$
- B$16 \times 10^{3} \text{ A/m}$
- C$4 \times 10^{3} \text{ A/m}$
- D$10^{3} \text{ A/m}$
Explore More
Similar Questions
$A$ toroid of $n$ turns,mean radius $R$ and cross-sectional radius $a$ carries current $I$. It is placed on a horizontal table taken as $xy$-plane. Its magnetic moment $m$ is:
The magnetic flux near the axis and inside the air core solenoid of length $60 \, cm$ carrying current '$I$' is $1.57 \times 10^{-6} \, Wb$. Its magnetic moment will be $[\mu_0 = 4 \pi \times 10^{-7} \, SI \, unit$ and cross-sectional area is very small as compared to the length of the solenoid.] (in $Am^2$)
$A$ winding wire which is used to frame a solenoid can bear a maximum $10\, A$ current. If the length of the solenoid is $80\, cm$ and its cross-sectional radius is $3\, cm$,then the required length of the winding wire is $(B = 0.2\, T)$.
Medium
View Solution$A$ long solenoid has $200$ turns per $cm$ and carries a current of $2.5 \ A$. The magnetic field at its centre is $(\mu_0 = 4\pi \times 10^{-7} \ \text{Wb/A} \cdot \text{m})$.
Easy
View SolutionThe current required to be passed through a solenoid of $15\,cm$ length and $60$ turns in order to demagnetize a bar magnet of magnetic intensity $2.4 \times 10^3\,A/m$ is $.........A$.
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo