Class 12PhysicsMoving Charges and MagnetismAmpere’s circuital law and its application (Solenoid and Toroid)
Easy$A$ solenoid of $1200$ turns is wound uniformly in a single layer on a glass tube $2\,m$ long and $0.2\,m$ in diameter. The magnetic intensity at the center of the solenoid when a current of $2\,A$ flows through it is:
- A$2.4 \times 10^3\,A m^{-1}$
- B$1.2 \times 10^3\,A m^{-1}$
- C$1\,A m^{-1}$
- D$2.4 \times 10^{-3}\,A m^{-1}$
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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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View SolutionThe relative permeability in a core of a solenoid is $400$. The windings of a solenoid are insulated from the core and carry a current of $2 \ A$. If the number of turns is $1000$ per meter,then the magnetic intensity inside the core of the solenoid is . . . . . . $\frac{A}{m}$.
$A$ long wire carrying a current of $18 \,A$ is kept along the axis of a long solenoid of radius $1 \,cm$. The magnetic field due to the solenoid is $8.0 \times 10^{-3} \,T$. The magnitude of the resultant magnetic field at a point $0.6 \,mm$ from the solenoid axis is (Assume $\mu_0 = 4 \pi \times 10^{-7} \,Tm/A$):
$A$ current $i$ $A$ flows along an infinitely long straight thin-walled tube. The magnetic induction at any point inside the tube is:
The 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 $.........A$.
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