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## A current I = 20 A is directed along the positive x-axis and perpendicular to a magnetic field. A magnetic force per unit length of 0.16 N/m

Question

A current I = 20 A is directed along the positive x-axis and perpendicular to a magnetic field. A magnetic force per unit length of 0.16 N/m acts on the conductor in the negative y-direction. Calculate the magnitude and direction of the magnetic field in the region through which the current passes. magnitude T direction

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Physics
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2021-09-04T10:39:59+00:00
2021-09-04T10:39:59+00:00 2 Answers
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## Answers ( )

Answer:the magnitude and direction of the magnetic field in the region through which the current passes is 0.008 T and +z direction.Explanation:given information:

current, I = 20 A

magnetic force per unit length, F/L = 0.16 N/m

the conductor in the negative y-direction

θ = 90° (perpendicular)

as we know the formula to calculate magnetic force is

F = B I L sin θ

B = F/(I L sin θ)

= (F/L) (1/I sin θ)

= 0.16 (1/15 sin 90)

=

0.008 Tsince F is in the negative y direction, based of the right hand rule the magnetic field is in positive z directionAnswer:

Explanation:

Given:

current, I = 20 A

Magnetic force per unit length, F/L

= 0.16 N/m

Conductor in the negative y-direction, therefore θ = 90° (perpendicular)

For a magnetic field,

F = B I L sin θ

B = F/(I L sin θ)

= 0.16 × (1/15 sin 90)

= 0.008 T

The field is in the +ve z – direction.