Chapter 20, Problem 73GP

To determine

The magnitude of current flowing and the direction of the magnetic field.

Answer to Problem 73GP

Current in the bar is $1.6\text{ A}$ . The direction of the magnetic field must be towards down.

Explanation of Solution

Given:

Length of the bar is $22\text{ cm}$ and mass of the bar is $1.5\text{ g}$ . The magnitude of magnetic field is $1.7\text{ T}$ . Bar is moving from the rest with the speed of $28\text{ m/s}$ in a distance of $1\text{ m}$

Formula used:

According to the third equation of motion,

  $v^2=u^2+2as$

Magnetic force is calculated as

  $F=IlB$

Calculation:

Acceleration due to the magnetic force is

  $\begin{array}{c}a=\frac{v^2-u^2}{2s}\\ =\frac{v^2-0}{2s}\end{array}$

Net force on the bar is,

  $\begin{array}{c}{F}_{net}=ma\\ =IlB\end{array}$

Thus, current on the bar is,

  $\begin{array}{c}I=\frac{ma}{lB}\\ =\frac{m\left(\frac{v^2}{2s}\right)}{lB}\\ \text{=}\frac{(1.5\times {10}^{-3}\text{ kg}){(28\text{ m/s})}^2}{2(1\text{ m})(0.22\text{ m})(1.7\text{ T})}\\ =1.6\text{ A}\end{array}$

Using the right-hand rule, the current is towards the expanding circuit, the magnetic force is thus in the downward direction.

Conclusion:

Current in the bar is $1.6\text{ A}$ and thus, the direction of the magnetic field must be towards down.