Question

A wire carrying a 25-A current passes between the poles of a strong magnet that is perpendicular to its field and experiences a 5.96-N force on the 3.86 cm of wire in the field. What is the average field strength?

Answer 1

The magnitude of the force on a wire of length l carrying a current I and making an angle $\theta$ with a magnetic field of density B given by :

$|\overrightarrow{AB}| = B.I.l.sin(\theta)$

Rearranging to get $\overrightarrow{B}$

$B = \dfrac{|\overrightarrow{F}| }{Ilsin\theta}\\ \space\space\space\space = \dfrac{5.96}{25\times(3.86\times10^{-2})\times sin \dfrac{\pi}{2}}\\ \space\space\space\space= \dfrac{5.96}{0.965}\\$

$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space = 6.167T$

answer B= 6.167T