Class 12PhysicsMoving Charges and MagnetismAmpere’s circuital law and its application (Solenoid and Toroid)
Medium$A$ closely wound solenoid of $80 \,cm$ long has $5$ layers of windings of $400$ turns each. The diameter of the solenoid is $1.8 \,cm$. If the current carried is $8 \,A$, then the magnitude of the magnetic field inside the solenoid near its centre is approximately
- A$1.5 \times 10^{-2} \,T$
- B$2.5 \times 10^{-2} \,T$
- C$3.5 \times 10^{-2} \,T$
- D$4.5 \times 10^{-2} \,T$
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$A$ long solenoid is formed by winding $20$ $turns/cm$. The current necessary to produce a magnetic field of $20$ $mT$ inside the solenoid will be approximately ..... $A$ $(\frac{\mu_0}{4\pi} = 10^{-7} \text{ T m/A})$.
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View Solution$A$ long solenoid has a radius $a$ and the number of turns per unit length is $n$. If it carries a current $i$,then the magnetic field on its axis is directly proportional to
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View SolutionThere are $50$ turns per $cm$ length in a very long solenoid. It carries a current of $2.5 \ A$. The magnetic field at its centre on the axis is . . . . . . $T$.
$A$ long solenoid is formed by winding $70$ turns $cm^{-1}$. If $2.0\,A$ current flows,then the magnetic field produced inside the solenoid is $.......\times 10^{-4}\,T$ $\left(\mu_0 = 4\pi \times 10^{-7}\,TmA^{-1}\right)$
$A$ toroid has an iron core with an internal magnetic field of $10 \pi \text{ mT}$,when the current in the winding of $1500 \text{ turns/m}$ is $10 \text{ A}$. Determine the field due to magnetization $(\mu_0 = 4 \pi \times 10^{-7} \text{ H m}^{-1})$.
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