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Standard XII

Physics

Lenz's Law

In a coil of ...

Question

In a coil of area $10$ $c m^{2}$ and $10$ turns with magnetic field directed perpendicular to the plane and is changing at the rate of $10^{4}$ gauss/ second. The resistance of the coil is $20$ $Ω$ . The current in the coil will be

0.5 A

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5 A

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50 A

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5 \times 10^{8} A

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Solution

The correct option is A $5 A$
Induced current in the coil will be
$i = \frac{e m f}{R} = \frac{d B}{d t} \times \frac{A \times n}{R}$
$= \frac{10^{4}}{10^{4}} \times \frac{10 \times 10}{20}$
$= 5$ $A .$

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A rectangular coil of $20$ turns and area of cross section $25 s q . c m$ has a resistance of $100 o h m$ . If a magnetic field which is perpendicular to the plane of the coil changes at a rate of $1000 t e s l a p e r s e c$ , the current in the coil is:

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Q. A coil of 1200 turns and mean area of 500

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In a coil of area 10 cm2 and 10 turns with magnetic field directed perpendicular to the plane and is changing at the rate of 104 gauss/ second. The resistance of the coil is 20 Ω . The current in the coil will be

The correct option is A 5AInduced current in the coil will be i=emfR=dBdt×A×nR =104104×10×1020 =5 A.

In a coil of area $10$ $c m^{2}$ and $10$ turns with magnetic field directed perpendicular to the plane and is changing at the rate of $10^{4}$ gauss/ second. The resistance of the coil is $20$ $Ω$ . The current in the coil will be

The correct option is A $5 A$
Induced current in the coil will be
$i = \frac{e m f}{R} = \frac{d B}{d t} \times \frac{A \times n}{R}$
$= \frac{10^{4}}{10^{4}} \times \frac{10 \times 10}{20}$
$= 5$ $A .$