Figure, shows an equilateral triangular loop CDE, carrying current I. Length of each side of triangle is l. If a uniform magnetic field exists parallel to side DE of loop, then find the foces acting on the three wires CD, DE and EC separately.

Answer

`IlBsqrt3//2`, normally outwards, zero `IlBsqrt3//2`, normally inwards

Explanation

For current in wire CD, the angle between `vecl` and `vecB` is `120^@`, i.e., `theta=120^@`. Therefore, force on wire CD is
`vecF_(CD)=I[(vec(CD))xxvecB]=IlBsin 120^@hatn`
or `F_(CD)=IlBsin120^@=IlBxxsqrt3//2`
Its direction according to Right Hand Rule is normally out of the plane of paper.
For arm DE, the anggle between `vecl` and `vecB` is zero i.e., `theta=0^@`. Therefore, force on wire DE is `IlBsin0^@=0`
For arm EC, the angle between `vecl` and `vecB` is `120^@` i.e., `theta=120^@`. Therefore, force on wire EC is `F_(CE)I(CE)Bsin120^@=IlBxxsqrt3//2`
Its direction is normally into the plane of paper.