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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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