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$10^{3} \text{ A/m}$
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Similar Questions
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 $Am^2$)
For a long current-carrying solenoid,the magnetic field inside is $0.6 \ T$. The magnetic energy per unit volume is . . . . . . .
$A$ solenoid of $1.5 \ m$ length and $4.0 \ cm$ diameter possesses $10 \ turns/cm$. $A$ current of $5 \ A$ is flowing through it. The magnetic induction at the axis inside the solenoid is:
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View Solution$A$ toroidal solenoid with air core has an average radius $R$,number of turns $N$,and area of cross-section $A$. The self-inductance of the solenoid is (Neglect the field variation across the cross-section of the toroid).
If a long hollow copper pipe carries a current,then the magnetic field is produced:
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