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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Advanced 2018 · Paper 2 · Physics · Magnetic Effects of Current

A moving coil galvanometer has 50 turns and each turn has an area 2...

A moving coil galvanometer has 50 turns and each turn has an area 2×10−4m22 \times {10}^{- 4}{m}^{2} . The magnetic field produced by the magnet inside the galvanometer is 0.02T0.02T . The torsional constant of the suspension wire is 10−4Nmrad−1{10}^{- 4}Nm{rad}^{- 1} . When a current flows through the galvanometer, a full scale deflection occurs if the coil rotates by 0.2rad0.2rad . The resistance of the coil of the galvanometer is 50Ω50Ω . This galvanometer is to be converted into an ammeter capable of measuring current in the range 0−1.0A0 - 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 __________.A moving coil galvanometer has 50 turns and each turn has an area 2 × 10 ^ - 4 m ^ 2 . The magnetic field produced by

Correct answer

0

Step-by-step solution

{n=50turnsA=2×10−4m2\begin{cases}n = 50turns & A = 2 \times {10}^{- 4}{m}^{2}\end{cases}
{B=0.02TK=10−4\begin{cases}B = 0.02T & K = {10}^{- 4}\end{cases}
{Qm=0.2radRg=50Ω\begin{cases}{Q}_{m} = 0.2rad & {R}_{g} = 50Ω\end{cases}
{IA=0−1.0Aτ=MB=Cθ,M=NIA\begin{cases}{I}_{A} = 0 - 1.0A & τ = MB = C\theta, M = NIA\end{cases}
BIgNA=CθB {I}_{g} NA = C\theta
0.02×Ig×50×2×10−4=10−4×0.20.02 \times {I}_{g} \times 50 \times 2 \times {10}^{- 4} = {10}^{- 4} \times 0.2
Ig=0.1A{I}_{g} = 0.1A
For galvanometer, resistance is to be connected to ammeter in shunt.
A moving coil galvanometer has 50 turns and each turn has an area 2 × 10 ^ - 4 m ^ 2 . The magnetic field produced by
Ig×Rg=(I−Ig)S{I}_{g} \times {R}_{g} = \left(I - {I}_{g}\right)S
0.1×50=(1−0.1)S0.1 \times 50 = \left(1 - 0.1\right)S
S=509=5.55S = \dfrac{50}{9} = 5.55
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