A uniform conducting wire A B C has a mass of 10 g. A current of 2 A flows through it. The wire is kept in a uniform magnetic field B=2 T. The acceleration of the wire will be

Question

A uniform conducting wire A B C has a mass of 10 g. A current of 2 A flows through it. The wire is kept in a uniform magnetic field B=2 T. The acceleration of the wire will be (A) Zero (B) 12 ms -2 (C) 1.2 × 10-3 ms -2 (D) 0.6 × 10-3 ms -2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0cb751469c3f32d16ed24db37a98adcb-.png" /> Force on wire A B C will be same as force on wire A C using F=i( vecI × vecB) F =2((3/100))(2) sin 90° =12 × 10-2 N a=(F/m) =(12 × 10-2/10-2)=12 m / s 2

Q. A uniform conducting wire $A BC$ has a mass of $10 g$ . A current of $2 A$ flows through it. The wire is kept in a uniform magnetic field $B = 2 T$ . The acceleration of the wire will be

Zero

$12 m s^{- 2}$

$1.2 \times 1 0^{- 3} m s^{- 2}$

$0.6 \times 1 0^{- 3} m s^{- 2}$

Force on wire $A BC$ will be same as force on wire $A C$
using $F = i (I \times B)$
$F = 2 (\frac{3}{100}) (2) sin 9 0^{\circ}$
$= 12 \times 1 0^{- 2} N$
$a = \frac{F}{m} = \frac{12 \times 1 0 ^{- 2}}{1 0 ^{- 2}} = 12 m / s^{2}$

Q. A uniform conducting wire ABC has a mass of 10g. A current of 2A flows through it. The wire is kept in a uniform magnetic field B=2T. The acceleration of the wire will be

Zero

12ms−2

1.2×10−3ms−2

0.6×10−3ms−2

Force on wire ABC will be same as force on wire AC using F=i(I×B) F=2(1003​)(2)sin90∘ =12×10−2N a=mF​=10−212×10−2​=12m/s2

Q. A uniform conducting wire $A BC$ has a mass of $10 g$ . A current of $2 A$ flows through it. The wire is kept in a uniform magnetic field $B = 2 T$ . The acceleration of the wire will be

$12 m s^{- 2}$

$1.2 \times 1 0^{- 3} m s^{- 2}$

$0.6 \times 1 0^{- 3} m s^{- 2}$

Force on wire $A BC$ will be same as force on wire $A C$
using $F = i (I \times B)$
$F = 2 (\frac{3}{100}) (2) sin 9 0^{\circ}$
$= 12 \times 1 0^{- 2} N$
$a = \frac{F}{m} = \frac{12 \times 1 0 ^{- 2}}{1 0 ^{- 2}} = 12 m / s^{2}$