Question

Determine the magnitude and direction of the force on a 200 m power line carrying a current of 5.0-A due west in a magnetic field of 6.0 μΤ in a direction of 30°, north of east.

Answer 1

Answer:

Explanation:

l = 200 m

i = 5 A west

B = 6 micro Tesla direction 30 degree north of east

angle between length vector and the magnetic field vector = 180 - 30 = 150 degree

Write the length and the magnetic field in the vector form

$\overrightarrow{l}=- 200 \widehat{i}metre$

$\overrightarrow{B}= 6\times 10^{-6}\left ( Cos30\widehat{i}+Sin30\widehat{j} \right )Tesla$

$\overrightarrow{B}=\left ( 5.2\widehat{i}+3\widehat{j} \right )\times 10^{-6}Tesla$

$\overrightarrow{F} = i \overrightarrow{l}\times \overrightarrow{B}$

$\overrightarrow{F} = 5\times \left ( -200\widehat{i} \right )\times \left ( 5.2\widehat{i}+3\widehat{j} \right )\times 10^{-6}$

$\overrightarrow{F} =- 3\times 10^{-3}\widehat{k}Newton$

Thus, the magnitude of force is 3 x 10^-3 newton and it is directed towards negative z axis direction.