The figure below shows an electrically conductive bar of square cross-section resting on a plane surface. The bar of mass of 1 kg has a depth of 0.5 m along the y direction. The coefficient of friction between the bar and the surface is 0.1. Assume the acceleration due to gravity to be 10 m/s\(^2\). The system faces a uniform flux density \( B = -1 \hat{z} \, \text{T} \). At time \( t = 0 \), a current of 10 A is switched onto the bar and is maintained. When the bar has moved by 1 m, its speed in metre per second is \(\underline{\hspace{2cm}}\) (rounded off to one decimal place).
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When calculating the motion of a bar in a magnetic field, account for both the magnetic force and the frictional force, and apply Newton's second law.
The induced force on the bar due to the magnetic field is given by:
\[
F = I L B
\]
where \( I = 10 \, \text{A} \), \( L = 1 \, \text{m} \) (length of the bar), and \( B = 1 \, \text{T} \). Thus:
\[
F = 10 \times 1 \times 1 = 10 \, \text{N}.
\]
The frictional force \( f \) is given by:
\[
f = \mu mg = 0.1 \times 1 \times 10 = 1 \, \text{N}.
\]
The net force is \( F - f = 10 - 1 = 9 \, \text{N} \). Using Newton's second law \( F = ma \), where \( m = 1 \, \text{kg} \):
\[
a = \frac{9}{1} = 9 \, \text{m/s}^2.
\]
Finally, using the kinematic equation \( v^2 = u^2 + 2as \), where \( u = 0 \), \( s = 1 \, \text{m} \), and \( a = 9 \, \text{m/s}^2 \):
\[
v^2 = 0 + 2 \times 9 \times 1 = 18 $\Rightarrow$ v = \sqrt{18} \approx 4.24 \, \text{m/s}.
\]
Thus, the speed is approximately \( 4.2 \, \text{m/s} \).
