JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 14)
A moving coil galvanometer has a coil with
175 turns and area 1 cm2. It uses a torsion band
of torsion constant 10–6 N-m/rad. The coil is
placed in a maganetic field B parallel to its
plane. The coil deflects by 1° for a current of
1 mA. The value of B (in Tesla) is
approximately :-
10–4
10–2
10–1
10–3
Explanation
NIAB = KQ
175 × 1 × 10–3 × 1 × 10–4 × B = $${{{{10}^{ - 6}} \times \pi } \over {180}}$$
$$ \Rightarrow B = {\pi \over {180}} \times {{10} \over {175}} \approx 9.97 \times {10^{ - 4}}\,T$$
$$ \Rightarrow $$ B = 10–3 T
175 × 1 × 10–3 × 1 × 10–4 × B = $${{{{10}^{ - 6}} \times \pi } \over {180}}$$
$$ \Rightarrow B = {\pi \over {180}} \times {{10} \over {175}} \approx 9.97 \times {10^{ - 4}}\,T$$
$$ \Rightarrow $$ B = 10–3 T
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