Aloop of wire carrying a current of 2.0 a is in the shape of a right triangle with two equal sides, each 15 cm long. a 0.7 t uniform magnetic field is in the plane of the triangle and is perpendicular to the hypotenuse. the resultant magnetic force on the two equal sides has a magnitude of

0 (zero )

Explanation:

We shall represent sides of the triangle as follows

one side 15 x 10⁻² i

Another side - 15 x 10⁻² j

Magnetic field can be represented as follows

B = - 0.7 cos 45 i - 0  .7 sin 45 j

= - .7 √2 i - .7 √2 j

Magnetic force on first side

F₁ = i ( L x B )

= 2  ( 15 x 10⁻² i  ) x (  - .7 √2 i - .7 √2 j  )

- 2 x 0 .7 √2 k

=  - 29.7  k

Magnetic force on second  side

F₂ = 2 (  - 15 x 10⁻² j ) x (  - .7 √2 i - .7 √2 j  )

29.7 k

Net force

=  - - 29.7  k +. - 29.7  k

= 0 (zero )

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this