Question

One coil of resistance 40 ohm is connected to a galvanometer of 160 ohm resistance

Answer 1

Total resistance
$40 + 160 \\ = 200 \: ohm$

Answer 2

Answer: ∆B = 0.566 T

Given: Resistances = 40 ohm, 160 ohm

n = 100 , r = 6mm , Q = 32 micro coulomb  

Solution:

Total resistance = 40ohm + 160 ohm = 200 ohm

n = 100, r = 6mm = $6 \times 10^{-3} m$

Q = 32 micro coulomb = $32 \times 10^{-6} coulomb$

Wkt, rate of change of magnetic field ∆B = ∆Q \times R/ A

Where, A = n(πr2)

              $A = 100 \times 3.14 \times (6 \times 10^{-3})^2$

              $A = 314 \times 36 \times 10^{-6} =11304 \times 10^{-6}$

Hence,

              $\Delta B = (32 \times 10^{-6} \times 200)/ 11304 \times 10^{-6}$

               ∆B = (6400)/ 11304

               ∆B = 0.566 T