Class 12PhysicsMoving Charges and MagnetismAmpere’s circuital law and its application (Solenoid and Toroid)
Easy$A$ long solenoid carrying a current produces a magnetic field $B$ along its axis. If the number of turns per cm are tripled and the current is made $\left(\frac{1}{4}\right)^{th}$,then the new value of the magnetic field will be:
- A$\frac{B}{3}$
- B$\frac{B}{4}$
- C$\frac{3 B}{4}$
- D$\frac{2 B}{3}$
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The magnetic field intensity $(H)$ at the centre of a long solenoid having '$n$' turns per unit length and carrying a current $I$, when no material is kept in it is ($\mu_0$ = permeability of free space)
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View Solution$A$ long solenoid has $200$ turns per $cm$ and carries a current $i$. The magnetic field at its centre is $6.28 \times 10^{-2} \ Wb/m^2$. Another long solenoid has $100$ turns per $cm$ and it carries a current $i/3$. The value of the magnetic field at its centre is:
$A$ 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 $(B = 0.2\, T)$.
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View Solution$A$ long solenoid with $20$ turns per $cm$ is made. To produce a magnetic field of $20$ $mT$ inside the solenoid, the necessary current will be approximately......$A$.
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