Question

heya in need of proper explanation with labelled diagram...​

Answer 1

According to the given question,

A smooth plane is inclined at an angle of 30 ° with the horizontal

A current carrying metallic rod is placed parallel to the ground

The plane is located in a uniform magnetic field

B = 0.15 T in vertical direction

Direction of force can be found out by doing

L × B

So the cos component of force will act along the plane of the rod opposite to the sin component of gravitaional force so that the rod remains stationary

$mg \sin(30) = F \cos(30)$

$mg \sin(30) =B i \: l \cos(30)$

sin 30 = 1/2

cos 30 = √3 /2 A

B = 0.15 T

m / l = 0.03 kg/ m

Substituting the values in the above equation ,

$i = \frac{mg \sin(30) }{l \times B \cos(30) }$

$i = \frac{0.03 \times 10 \times 1 \times 2}{2 \times 0.15 \times \sqrt{3} }$

$i = \frac{0.3}{0.26}$

$\boxed{i = 1.15A}$

I hope it's correct