Question

A potential difference of 5.0×10^-2 V is developed across a 10-cm-long wire as it moves through a magnetic field at 5.0 m/s. The magnetic field is perpendicular to the axis of the wire.

What is the strength and direction of the magnetic field?

 

Please note: The magnetic field is perpendicular to BOTH the axis of the piece of wire AND to the DIRECTION OF MOTION.

Answer 1

Concepts and reason

The concepts required to solve the given question is strength and direction of magnetic field.

Initially, convert the length from cm to m. Later, write and expression showing the relation between the induced EMF and the magnetic field. Finally, calculate the strength of magnetic field.

Fundamentals

The expression for the induced EMF is as follows:

$\varepsilon = Blv$

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

Rearrange this for the value of B.

$B = \frac{\varepsilon }{{lv}}$

The length is as follows:

$l = 10{\rm{ cm}}$

Convert to m as follows:

$\begin{array}{c}\\l = 10{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1.00{\rm{ cm}}}}} \right)\\\\ = 0.10{\rm{ m}}\\\end{array}$

Substitute $5.0 \times {10^2}{\rm{ V}}$ for $\varepsilon$, 0.10 m for l, and 5.0 m/s for v in the equation $B = \frac{\varepsilon }{{lv}}$.

$\begin{array}{c}\\B = \frac{{5.0 \times {{10}^2}{\rm{ V}}}}{{\left( {0.10{\rm{ m}}} \right)\left( {5.0{\rm{ m/s}}} \right)}}\\\\ = 0.10{\rm{ T}}\\\end{array}$

Ans:

The strength of the magnetic field is equal to 0.10 T.