Question

A uniform, 458 g metal bar 77.5 cm long carries a current I in a uniform, horizontal 1.25 T magnetic field as shown in the following figure. The directions of I and B⃗ are shown in the figure. The bar is free to rotate about a friction less hinge at point b. The other end of the bar rests on a conducting support at point a but is not attached there. The bar rests at an angle 60.0 ∘ of above the horizontal.(Figure 1)

Answer 1

Answer:

$21 { \frac{?}{?} }^{?}$

Answer 2

The current is 1.71A.

Mass of the metal bar = 458 g  (Given)

Length of the metal bar = 77.5 cm  (Given)

Magnetic field  = 1.25 T (Given)

According to the formula of magnetic force -

Fb = mgcos60

BIL = mgcos60

I = mgcos60/BL

= 458 × 10³ × 9.8cos60/ 1.25 × 77.5 × 10³

= 1.71

Therefore, the current is 1.71A.