Class 12PhysicsMoving Charges and MagnetismAmpere’s circuital law and its application (Solenoid and Toroid)
EasyThe magnetic intensity at the centre of a long current-carrying solenoid is found to be $1.6 \times 10^3 \text{ A m}^{-1}$. If the number of turns is $8 \text{ per cm}$,then the current flowing through the solenoid is $................\, \text{A}$.
- A$9$
- B$2$
- C$5$
- D$6$
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Similar Questions
$A$ long solenoid carrying a current produces a magnetic field $B$ along its axis. If the number of turns per $cm$ is doubled and the current is made $\left(\frac{1}{3}\right)^{rd}$ of its original value,then the new value of the magnetic field will be:
$A$ solenoid is $1.0 \ m$ long and it has $4250$ turns. If a current of $5.0 \ A$ is flowing through it, what is the magnetic field at its centre? $[\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A]$
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View SolutionGive the formula for the magnetic field due to a toroid.
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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 $mT$)
$A$ toroid of $n$ turns,mean radius $R$ and cross-sectional radius $a$ carries current $I$. It is placed on a horizontal table taken as $xy$-plane. Its magnetic moment $m$ is:
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