A moving coil galvanometer has 48 turns and area of coil is 4× 10-2m2 . If the magnetic field is 0.2 T, then to increase the current sensitivity by 25% without changing area (A) and field (B) the number of turns should become

Question

A moving coil galvanometer has 48 turns and area of coil is 4× 10-2m2 . If the magnetic field is 0.2 T, then to increase the current sensitivity by 25% without changing area (A) and field (B) the number of turns should become (A) 24 (B) 36 (C) 60 (D) 54

Tardigrade Admin · Accepted Answer

As we know Current sensitivity Si=(NBA/C) ⇒ Si propto N⇒ ((Si)1/(Si)2)=(N1/N2) ⇒ (100/125)=(48/N2)⇒ N2=60

Q. A moving coil galvanometer has 48 turns and area of coil is $4 \times 10^{- 2} m^{2}$ . If the magnetic field is 0.2 T, then to increase the current sensitivity by 25% without changing area (A) and field (B) the number of turns should become

24

36

60

54

As we know Current sensitivity $S_{i} = \frac{NB A}{C}$ $\Rightarrow$ $S_{i} \propto N \Rightarrow \frac{( S _{i} ) _{1}}{( S _{i} ) _{2}} = \frac{N _{1}}{N _{2}}$ $\Rightarrow$ $\frac{100}{125} = \frac{48}{N _{2}} \Rightarrow N_{2} = 60$

Q. A moving coil galvanometer has 48 turns and area of coil is 4×10−2m2 . If the magnetic field is 0.2 T, then to increase the current sensitivity by 25% without changing area (A) and field (B) the number of turns should become

24

36

60

54

As we know Current sensitivity Si​=CNBA​ ⇒ Si​∝N⇒(Si​)2​(Si​)1​​=N2​N1​​ ⇒ 125100​=N2​48​⇒N2​=60

Q. A moving coil galvanometer has 48 turns and area of coil is $4 \times 10^{- 2} m^{2}$ . If the magnetic field is 0.2 T, then to increase the current sensitivity by 25% without changing area (A) and field (B) the number of turns should become

As we know Current sensitivity $S_{i} = \frac{NB A}{C}$ $\Rightarrow$ $S_{i} \propto N \Rightarrow \frac{( S _{i} ) _{1}}{( S _{i} ) _{2}} = \frac{N _{1}}{N _{2}}$ $\Rightarrow$ $\frac{100}{125} = \frac{48}{N _{2}} \Rightarrow N_{2} = 60$