As shown in the figure, a metallic rod of linear density 0.45 kg m -1 is lying horizontally on a smooth incline plane which makes an angle of 45° with the horizontal. The minimum current flowing in the rod required to keep it stationary, when 0.15 T magnetic field is acting on it in the vertical upward direction, will be: . Use .g=10 m / s 2

Question

As shown in the figure, a metallic rod of linear density 0.45 kg m -1 is lying horizontally on a smooth incline plane which makes an angle of 45° with the horizontal. The minimum current flowing in the rod required to keep it stationary, when 0.15 T magnetic field is acting on it in the vertical upward direction, will be: . Use .g=10 m / s 2  (A) 30 A (B) 15 A (C) 10 A (D) 3 A

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/11ff6d10ad751b8b25eb029faa31bf00-.png" /> mg sin 45°= ILB cos 45° ∴ I =(( m / L )) ( g / B ) =((0.45)(10)/0.15)=30 A

$30 A$

$15 A$

$10 A$

$3 A$

$m g sin 4 5^{\circ} = I L B cos 4 5^{\circ}$
$∴ I = (\frac{m}{L}) \frac{g}{B}$
$= \frac{( 0.45 ) ( 10 )}{0.15} = 30 A$

30A

15A

10A

3A

mgsin45∘=ILBcos45∘ ∴I=(Lm​)Bg​ =0.15(0.45)(10)​=30A

$30 A$

$15 A$

$10 A$

$3 A$

$m g sin 4 5^{\circ} = I L B cos 4 5^{\circ}$
$∴ I = (\frac{m}{L}) \frac{g}{B}$
$= \frac{( 0.45 ) ( 10 )}{0.15} = 30 A$