Class 12PhysicsMoving Charges and MagnetismAmpere’s circuital law and its application (Solenoid and Toroid)
Easy$A$ solenoid has a core of a material with relative permeability $400$. The windings of the solenoid are insulated from the core and carry a current of $1 \text{ A}$. If the number of turns is $1000 \text{ per metre}$,the magnetic field $(B)$ is . . . . . . $\text{T}$. (Given: $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$)
- A$1.6\pi \times 10^2$
- B$16\pi \times 10^2$
- C$16\pi \times 10^{-2}$
- D$0.16\pi \times 10^{-2}$
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Similar Questions
$A$ long solenoid produces a magnetic field $B$ on its axis. If it is cut into four equal parts and,for the same current,half the number of turns are wound on any one piece,the value of the magnetic field on its axis becomes:
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View Solution$A$ toroid has a non-ferromagnetic core of inner radius $r_{1}$ and outer radius $r_{2}$,around which $N$ turns of wire are wound. If the current in the wire is $I$,then the magnetic field inside the toroid is ($\mu_{0} =$ permeability of free space).
The magnetic field $B$ within a solenoid having $n$ turns per meter length and carrying a current of $i$ amperes is given by:
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View Solution$A$ solenoid of $1 \ m$ length and $3.55 \ cm$ inner diameter carries a current of $5 \ A$. If the solenoid consists of five closely packed layers each with $700$ turns along its length,then the magnetic field at its centre is (in $mT$)
$A$ current $i$ flows in an infinitely long, straight and thin-walled pipe. Then,
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