Question

Six identical vertical metal bars start at the positions shownbelow and move at constant velocities through identical magneticfields. The bars make electrical contact with and move alongfrictionless metal rods attached to light bulbs.

At the instant shown, rank these six scenarios on the basis of themagnitude of the current in the light bulb.

Rank from largest to smallest. To rank items as equivalent, overlapthem.

Answer 1

Concepts and reason

The main concept required to solve this problem are the induced emf, magnetic field, length, velocity, current and resistance.

Initially, write the equation for the induced emf in the vertical bar, the equation for the current due to the induced emf. Use these equations and rank the currents from the largest to smallest.

Fundamentals

The equation for the induced emf in the vertical metal bar is,

$\varepsilon = Blv$

Here, B is the magnetic field, l is the length of the bar, and v is the speed of the bar.

The equation for the current due to the induced emf is,

$i = \frac{\varepsilon }{R}$

Here, $\varepsilon$ is the induced emf and R is the resistance.

The equation for the induced emf in the vertical metal bar is,

$\varepsilon = Blv$ …… (1)

Here, B is the magnetic field, l is the length of the metal bar and v is the vertical metal bar.

The equation for the current through the bulbs due to the induced emf is,

$i = \frac{\varepsilon }{R}$ …… (2)

Here, $\varepsilon$ is the induced emf and $R$ is the resistance in the bulb.

Substitute the equation (1) in above equation (2).

$\begin{array}{l}\\i = \frac{{\left( {Blv} \right)}}{R}\\\\i = \left( {\frac{{Bl}}{R}} \right)v\\\end{array}$

In the above equation $\frac{{Bl}}{R}$ is constant, therefore, the above equation can be written as follows,

$i\,\,\,\alpha \,\,\,v$

The equation for the relation between the current passing through the bulb and the velocity of the metal bar, that derived in above step1 is,

$i\,\,\,\alpha \,\,\,v$

Here, the current passing through the bulb is directly proportional to the velocity of the metal bar. If the velocity of the metal bar is more, the current passing through the bulb is more, if the velocity of the metal bar is less; the current passing through the bulb is less.

Therefore, the rank of the magnitude of the currents through the bulbs is as follows,

${i_{\left( {20{\rm{ cm/s}}} \right)}}\, > \,{i_{\left( {10{\rm{ cm/s}}} \right)}}\left( {{\rm{for both directions}}} \right) > {i_{\left( {{\rm{5 cm/s}}} \right)}} > {i_{\left( {0{\rm{ cm/s}}} \right)}}$

Ans:

${i_{\left( {20{\rm{ cm/s}}} \right)}}\, > \,{i_{\left( {10{\rm{ cm/s}}} \right)}}\left( {{\rm{for both directions}}} \right) > {i_{\left( {{\rm{5 cm/s}}} \right)}} > {i_{\left( {0{\rm{ cm/s}}} \right)}}$