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Question
Five long wires A, B, C, D and E, each carrying current I are arranged to form edges of a pentagonal prism as shown in figure. Each carries current out of the plane of paper.
- What will be magnetic induction at a point on the axis O? AxisE is at a distance R from each wire.
- What will be the field if current in one of the wires (say A) is switched off?
- What if current in one of the wire (say) A is reversed?
Solution
a. The wires shown in this problem carrying current outwards to the plane. And we know that direction of magnetic field is perpendicular to both current and position vector r. So, the vector sum of magnetic field produced by each wire at O is equal to 0.
Suppose the five wires A, B, C, D and E be perpendicular to the plane of paper at locations as shown in figure.
Thus, magnetic field induction due to five wires will be represented by various sides of closed pentagon in one order, lying in the plane of paper. So, its value is zero.
b. When current in AA' is switched off, then B1 = 0 and resultant becomes
R = B2 + B3 + B4 + B5
But form (a) part B1 + B2 + B3 + B4 + B5 = 0
Or `vecB_2 + vecB_3 + vecB_4 + vecB_5 = - vecB_1`
R = – B1
R = `(mu_0I)/(2pir)`
i.e. Direction of resultant is opposite to `vecB_1`.
c. If current in wire A is reversed, then
Total magnetic field induction at O = Magnetic field induction due to A + Magnetic field induction due to wires B, C, D and E
= `(mu_0)/(4piR) (2I)/R` (acting perpendicular to AO towards left) + `(mu_0)/pi (2I)/R` (acting perpendicular AO towards left)
= `(mu_0I)/(piR)` acting perpendicular AO towards left.
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