Q. A wire carrying a current of 3A is bent in the form of a parabola y^{2} = 4-x as shown in figure, where x and y are in metre. The wire is placed in a uniform magnetic field $\displaystyle \vec{B} = 5 \hat{k}$ tesla . The force acting on the wire is

(a) $ \displaystyle 60 \hat{i} N $

(b) $ \displaystyle -60 \hat{i} N $

(c) $\displaystyle 30 \hat{i} N $

(d) $ \displaystyle -30 \hat{i} N $

Ans: (a)

Sol: y^{2} = 4 – x

At x =0 , y = ± 2

$ \displaystyle \vec{L} = 4 \hat{j} $

(As , force on parabola is same as force on wire of length L )

$ \displaystyle \vec{F} = i (\vec{L} \times \vec{B}) $

$ \displaystyle \vec{F} = 3 (4 \hat{j} \times 5 \hat{k}) $

$ \displaystyle \vec{F} = 60 \hat{i} $