Question

A single-turn current loop, carrying a current of 4.05 A, is in the shape of a right triangle with sides 50.0, 120, and 130 cm. The loop is in a uniform magnetic field of magnitude 85.0 mT whose direction is parallel to the current in the 130 cm side of the loop. What are the magnitude of the magnetic forces on each of the three sides

Answer 1

`F_(130) = 0 N`

`F_(50) = 0.15889 N`

`F_(120)=0.15889 \ N`

Step-by-step explanation:

From the diagram below:

`B_x = - Bcos \theta` and
`B_y = Bsin \theta`

`tan \theta = (50)/(120)`

`\theta = tan^(-1) ((50)/(120))`

`\theta = 22.62 ^0`

Magnetic Force on the 130 cm length is
`\theta`:

`F = BILsin \theta \\\\`

`F = BILsin (0 )` ( since
`\theta` = 0 ; i.e parallel)

`F_(130) = 0 N`

Magnetic Force on the 50 cm length is:

`F_(50) = Il_yB_xCos \theta`

`F_(50) = 4.05*50*10^(-2)*85*10^(-3)cos (22.62)`

`F_(50) = 0.15889 N`

Magnetic Force on the 120 cm length is:

`F_(120)= IL_xB_yCos \theta`

`F_(120)= IL_xBsin \theta`

`F_(120)=4.05 * 120*10^(-2)*85*10^(-3)sin(22.62)`

`F_(120)=0.15889 \ N`