Question

A square metal wire loop of side 10 cm and resistance 1 Ω is moved with a constant velocity v in a uniform magnetic field of induction B = 2 Wbm⁻² as shown in figure. The magnetic field is perpendicular to the plane of the loop and directed into the paper. The loop is connected to a network of resistors each of value 3 Ω. With what speed should the loop be moved so that a steady current of 1 mA flows in the loop. [Ans: 2*10⁻² ms⁻¹]

Answer 1

first of all we have to find equivalent resistance of circuit. it seems like WEATSTONE BRIDGE.
so, Internel wire of resistance 3ohm is removed.
now, $\frac{1}{R'}=\frac{1}{3+3}\frac{1}{3+3}$
1/R' = 1/6 + 1/6 = 2/6
R' = 6/2 = 3ohm
R' and 1 Ω are joined in series
so, Req = R' + 1Ω = 3Ω + 1Ω = 4Ω

Let the speed of loop = v
current flows in the loop, i = 1mA = 0.001A
magnetic field , B = 2 Wb/m²
square metal wire loop of side , l = 10cm

use formula ,
emf , ξ = Bvl
we know from ohm's law , ξ = iR
Bvl = iR
2wb/m² × v × 10cm = 0.001A × 4Ω
2 × v × 0.1 = 0.001 × 4
v = 2 × 0.01 = 0.02 m/s

hence,speed of loop is 2 × 10^-2 m/s

Answer 2

Answer:

Explanation:

first of all we have to find equivalent resistance of circuit. it seems like WEATSTONE BRIDGE.

so, Internel wire of resistance 3ohm is removed.

now,

1/R' = 1/6 + 1/6 = 2/6

R' = 6/2 = 3ohm

R' and 1 Ω are joined in series

so, Req = R' + 1Ω = 3Ω + 1Ω = 4Ω

Let the speed of loop = v

current flows in the loop, i = 1mA = 0.001A

magnetic field , B = 2 Wb/m²

square metal wire loop of side , l = 10cm

use formula ,

emf , ξ = Bvl

we know from ohm's law , ξ = iR

Bvl = iR

2wb/m² × v × 10cm = 0.001A × 4Ω

2 × v × 0.1 = 0.001 × 4

v = 2 × 0.01 = 0.02 m/s