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Magnetism question

2018 · Shift 2 · Q44
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Magnetism question

2018 · Shift 2 · Q44

JEE AdvancedPhysicsMagnetismNumerical+3 / −1
A moving coil galvanometer has 505050 turns and each turn has an area 2×10−4 m2.2 \times {10^{ - 4}}\,{m^2}.2×10−4m2. The magnetic field produced by the magnet inside the galvanometer is 0.02T.0.02T.0.02T. The torsional constant of the suspension wire is 10−4 N m rad−1.{10^{ - 4}}\,N\,m\,ra{d^{ - 1}}.10−4Nmrad−1. When a current flows through the galvanometer, a full scale deflection occurs if the coil rotates by 0.2rad0.2rad0.2rad. The resistance of the coil of the galvanometer is 50Ω.50\Omega .50Ω. This galvanometer is to be converted into an ammeter capable of measuring current in the range 0−1.0A0-1.0A0−1.0A. For this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance, in ohms, is ‾\underline{\hspace{2cm}}​.
Numerical answer
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