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

Step-by-step 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* 10^(-6)\left ( Cos30\widehat{i}+Sin30\widehat{j} \right )Tesla`

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

`\overrightarrow{F} = i \overrightarrow{l}* \overrightarrow{B}`

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

`\overrightarrow{F} =- 3* 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.