91Ó°ÊÓ

The given data can be listed below as

The escape speed for the earth is, 11.2 km/s.

The magnetic field is, B = 0.80 T.

The value of current is, $I=2.0×{10}^{3}A$.

The value of mass is, m = 25kg.

The length of bar is, L = 50 cm.

The term magnetic field may be defined as the area around the magnet behave like a magnet.

The current and magnetic field perpendicular to each other so the force act on bar is

F = lLB …(1)

Hence, the magnitude is F = lLB and direction of the net force on the conducting bar is to right.

Using the equation of motion of determine the distance

$v^2=v_0^2+2a\left(x−x_0\right)$

The initial velocity is zero then,

$\begin{array}{r}{v}^2=2ad\\ d=\frac{v^2}{2a}\end{array}$

…(2)

And

F = ma

So

$\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}$ …(3)

From equation (1) and (3)

$a=\frac{lLB}{m}$ …(4)

From equation (2) and (4) than

$d=\frac{v^{2}m}{2ILB}$

Hence, the distance$d=\frac{v^{2}m}{2ILB}$.

The acceleration can be calculated as

$a=\frac{lLB}{m}$

Substitute all the value in the above equation.

$\begin{array}{r}\mathrm{a}=\frac{\left(2×{10}^3\mathrm{A}\right)(0.50\mathrm{m})(0.80\mathrm{T})}{25\mathrm{kg}}\\ \mathrm{a}=32\mathrm{m}/{\mathrm{s}}^2\end{array}$

Hence, the acceleration is$\mathrm{a}=32\mathrm{m}/{\mathrm{s}}^2$.

From equation (5) the distance of bar must travel along the rails is,

$d=\frac{v^{2}m}{2ILB}$

Substitute all the value in the above equation

$\begin{array}{r}\mathrm{d}=\frac{{\left(11.2×{10}^3\mathrm{m}/\mathrm{s}\right)}^2×25\mathrm{kg}}{2×2000\mathrm{A}×0.50\mathrm{m}×0.80\mathrm{T}}\\ =1.96×{10}^6\mathrm{m}\\ \mathrm{d}=1960\mathrm{km}\end{array}$

Hence the distance of bar must travel along the rails is, 1960 km.