A square loop of side 20 cm and resistance 1 Ω is moved towards right with a constant speed v 0. The right arm of the loop is in a uniform magnetic field of 5 T. The field is perpendicular to the plane of the loop and is going into it. The loop is connected to a network of resistors each of value 4 Ω. What should be the value of v 0 so that a steady current of 2 mA flows in the loop?

Question

A square loop of side 20 cm and resistance 1 Ω is moved towards right with a constant speed v 0. The right arm of the loop is in a uniform magnetic field of 5 T. The field is perpendicular to the plane of the loop and is going into it. The loop is connected to a network of resistors each of value 4 Ω. What should be the value of v 0 so that a steady current of 2 mA flows in the loop? (A) 1 m / s (B) 1 cm / s (C) 102 m / s (D) 10-2 cm / s

Tardigrade Admin · Accepted Answer

Equivalent circuit <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/3743d902026033a2807bb06e1e1ca9ed-.png" /> i =( V 0 B ℓ/4+1) ⇒ V 0=(5(2 mA )/5 × .2) =10-2 m / s =1 cm / s

$1 m / s$

$1 c m / s$

$1 0^{2} m / s$

$1 0^{- 2} c m / s$

Equivalent circuit

$i = \frac{V _{0} B ℓ}{4 + 1}$
$\Rightarrow V_{0} = \frac{5 ( 2 m A )}{5 \times .2}$
$= 1 0^{- 2} m / s = 1 c m / s$

1m/s

1cm/s

102m/s

10−2cm/s

Equivalent circuit i=4+1V0​Bℓ​ ⇒V0​=5×.25(2mA)​ =10−2m/s=1cm/s

$1 m / s$

$1 c m / s$

$1 0^{2} m / s$

$1 0^{- 2} c m / s$

Equivalent circuit

$i = \frac{V _{0} B ℓ}{4 + 1}$
$\Rightarrow V_{0} = \frac{5 ( 2 m A )}{5 \times .2}$
$= 1 0^{- 2} m / s = 1 c m / s$