Gujarati
5.Magnetism and Matter
easy

The number of turns and radius of cross-section of the coil of a tangent galvanometer are doubled. The reduction factor $K $ will be

A

$K$

B

$2K$

C

$4K$

D

$K/4$

Solution

(a)$K = \frac{{2R{B_H}}}{{{\mu _0}N}}$ ($R =$ radius, $N=$ number of turns)

Standard 12
Physics

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