Question

A tangent galvanometer has a coil of 25 turns and radius of 15 cm. The horizontal component of the earth’s magnetic field is 3 × 10^–5 T. The current required to produce a deflection of 45^o in it, is

• A. 0.29 A
• B. 1.2 A
• C. 3.6 × 10^–5 A
• D. 0.14 A

Answer 1

Answer: (A)

0.29 A

Answer 2

$i = \frac { 2 r B _ { H } } { \mu _ { 0 } N } \tan \theta$

$\Rightarrow i = \frac { 2 \times 15 \times 10 ^ { - 2 } \times 3 \times 10 ^ { - 5 } } { 4 \pi \times 10 ^ { - 7 } \times 25 } \times \tan 45 ^ { \circ }$ $\Rightarrow i = 0.29 \mathrm {~A}$