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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The magnetic field intensity $(H)$ at the centre of a long solenoid carrying a current of $2 \ A$ is found to be $1000 \ A/m$. The number of turns per centimeter of the solenoid is: (Use $\mu_0 = 4 \pi \times 10^{-7} \ T \ m \ A^{-1}$)
$A$ current $i$ is uniformly distributed over the cross section of a long hollow cylindrical wire of inner radius $R_1$ and outer radius $R_2$. The magnetic field $B$ varies with distance $r$ from the axis of the cylinder as:
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View Solution$A$ closely wound solenoid of $1200$ turns and area of cross-section $5 \,cm^2$ carries a current. If the magnetic moment of the solenoid is $1.2 \,J \,T^{-1}$, then the current through the solenoid is (in $\,A$)
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