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straight parallel wires A, B and C carryng current I=10 amp are on the corners of...

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kibauu | eNoter

Posted July 2, 2013 at 6:40 AM via web

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straight parallel wires A, B and C carryng current I=10 amp are on the corners of equilateral triangle with side a=30cm. The wires A carries current I into paper and wires B and C carry I out of the paper.

 

1. what is the direction of the total magnetic field on the center of gravity of the triangle?

2. calculate the magnitude of the total magnetic field

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted July 2, 2013 at 8:13 AM (Answer #1)

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The magnetic field around a wire due to the current flowing through the wire is equal to B = `(4*pi*10^-7*I)/(2*pi*r)` where I is the magnitude of current flowing in the wire and r is the distance of the point from the wire where the magnetic field is being measured.

The direction of field is given by the right hand rule. If the wire is held in the right hand such that current flows towards the thumbs the fingers curl in the direction of the induced magnetic field.

In the figure, all the three wires A, B and C carry an equal magnitude of current. The magnetic field induced due to all the three wires at the point O that is equidistant from the points A, B and C is the same. The field generated due to the flow of current in wires B and C is opposite in direction to that due to the current flowing in wire A. This gives the net magnetic field at point O as:

`(4*pi*10^-7*10)/(2*pi*(0.3/sqrt 3))`

= `(2*10^-7*10*sqrt 3)/(0.3)`

`~~ 1.15*10^-5` T

The direction of the field is in the anti-clockwise direction.

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llltkl | College Teacher | Valedictorian

Posted July 2, 2013 at 10:48 AM (Answer #2)

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Distance, r of each vertex from the centre of gravity of the equilateral triangle is given by

`r=2/3*sqrt3/2*a`

`=2/3*sqrt3/2*30` cm

`= 10sqrt3 ` cm

`=17.32` cm (approx.)

Magnetic field at the centre of gravity due to A,

`F_A =(mu_0*I)/(2*pi*r) = (4*pi*10-7 T m/A)*(10A)/(2pi*(0.1732 m))`

`= 1.1547*10^-5 T`

Magnetic field at the centre of gravity due to B and C will be the resultant of two vectors of magnitude F, operating at an angle of 120°

By the law of parallelogram of vectors, resultant of magnetic fields due to B and C at the centre of gravity,

`R_BC = sqrt(F^2+F^2+2F*Fcos120°)`

`=sqrt(2F^2-F^2)`

`=F`

As the direction of current in wire A is opposite to that in B and C, the two vectors `R_(BC)` and `F_A` will have directions that are reinforcing each other (see the attached image).

Thus overall magnetic field at the centre of gravity

`=F+F=2F= 2*1.1547*10^-5 T`

`=2.31 *10^-5 T` .

The direction of the overall magnetic field is perpendicular to the wires, along `F_A` with the anticlockwise magnetic lines of force.

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