JEE Main 2019 — Physics solution

JEE Main2019Physics

A moving coil galvanometer has a coil with 175 turns and area 1 cm². It uses a torsion band of torsion constant 10^–6N-m/rad. The coil is placed in a magnetic 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.

A
10^–4
B
10^–2
C
10^–1
D
10^–3

Solution

$N I A B = K Q 175 \times 1 \times 1 0^{- 3} \times 1 \times 1 0^{- 4} \times B = \frac{10 ^{- 6} \times π}{180} \Rightarrow B = \frac{π}{180} \times \frac{10}{175} \approx 9.97 \times 1 0^{- 4} T . \Rightarrow B = 1 0^{- 3} T .$

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