A square metal loop of side 10 cm and resistance 1Ω is moved with a constant velocity partly inside a uniform magnetic field of 2 Wbm-2, directed into the paper, as shown in the figure.

Question

A square metal loop of side 10 cm and resistance 1Ω is moved with a constant velocity partly inside a uniform magnetic field of 2 Wbm-2, directed into the paper, as shown in the figure. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/dd3b3a389697b6d20b3ec1f31db03fad-.png" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

ε = Bℓ v = 2 ×10-1 × v =0.2v I = (ε/R) =10-3 ⇒ (0.2v/4) = 10-3 [Since effective resistance R of bridge is R = (6 ×6/6 +6) = 3Ω so total resistance = 1 +3 =4Ω] ⇒ v =2cm s-1

Q. A square metal loop of side $10$ cm and resistance $1Ω$ is moved with a constant velocity partly inside a uniform magnetic field of $2 Wb m^{- 2}$ , directed into the paper, as shown in the figure.

$ε = B ℓ v = 2 \times 1 0^{- 1} \times v = 0.2 v$
$I = \frac{ε}{R} = 1 0^{- 3}$
$\Rightarrow \frac{0.2 v}{4} = 1 0^{- 3}$
[Since effective resistance $R$ of bridge is $R = \frac{6 \times 6}{6 + 6} = 3Ω$
so total resistance $= 1 + 3 = 4Ω$ ]
$\Rightarrow v = 2 c m s^{- 1}$

Q. A square metal loop of side 10 cm and resistance 1Ω is moved with a constant velocity partly inside a uniform magnetic field of 2Wbm−2, directed into the paper, as shown in the figure.

ε=Bℓv=2×10−1×v=0.2v I=Rε​=10−3 ⇒40.2v​=10−3 [Since effective resistance R of bridge is R=6+66×6​=3Ω so total resistance =1+3=4Ω] ⇒v=2cms−1

Q. A square metal loop of side $10$ cm and resistance $1Ω$ is moved with a constant velocity partly inside a uniform magnetic field of $2 Wb m^{- 2}$ , directed into the paper, as shown in the figure.

$ε = B ℓ v = 2 \times 1 0^{- 1} \times v = 0.2 v$
$I = \frac{ε}{R} = 1 0^{- 3}$
$\Rightarrow \frac{0.2 v}{4} = 1 0^{- 3}$
[Since effective resistance $R$ of bridge is $R = \frac{6 \times 6}{6 + 6} = 3Ω$
so total resistance $= 1 + 3 = 4Ω$ ]
$\Rightarrow v = 2 c m s^{- 1}$