Five long wires A, B, C, D and E, each carrying currentl I are arranged to form edges of a pentagonal prism as shown in figure. Each carries current out of the plane of paper.

a) What will be magnetic induction at a point on the axis O? Axis is at a distance R from each wire. b) What will be the field if current in one of the wires (say A) is switched off? c) What is current in one of the wire (say) A is reversed?

Options

A. `0, (mu_0 I)/(2pi R) , (mu_0 I)/(2pi R) ,(mu_0 I)/(pi R)`
B. `(mu_0I)/(2pi R) , 0 , (mu_0 I)/(pi R)`
C. `(mu_0 I)/(2pi R) , (mu_0 I)/(pi R) , (mu_0 I)/(pi R)`
D. `(mu_0 I)/(2pi R), (mu_0 I)/(2piR) , (mu_0I)/(pi R)`

Answer

A

Explanation

In figure `vecB_A, vecB_B , vecB_C , vecB_D " and " vecB_E` represent magnetic fields due to long wires A, B, C, D and E(each carrying current I out of the plane of paper) at a point on the axis O.
Note that `vecB_A bot OA, vecB_B bot OB, vecB_C bot OC, vecB_D bot OD, vecB_E bot OE`
a) Magnetic induction at O due to all five current carrying conductors, i.e
`vecB_R = vecB_A + vecB_B + vecB_C + vecB_D + vecB_E`
` = vecB_A + (vecB_C + vecB_D) (because vecB_B = - vecB_E)`
` = vecB_A - vecB_A ( because vecB_C + vecB_D = vecB_A) = 0`
b) Magnetic field at O due to four current carrying conductors B, C `vecB_R = vecB_B + vecB_C + vecB_D + vecB_E = (vecB_C + vecB_D) (because vecB_B = - vecB_E)`
` = 0 vecB_A (as vecB_C + vecB_D = - vecB_A)`
or `|vecB_A| = B_A = (mu_0I)/(2pi R)`
Perpendicular to AO, towards left
C) Magnetic field at O due to five current carrying conductors with current in wire A reversed, i.e
`B_R = - vecB_A + vecB_B + vecB_C + vecB_D + vecB_E`
` = -vecB_A + (vecB_C + vecB_D) (as vecB_B = - vecB_E)`
`= -vecB_A - vecB_A = - 2vecB_A`
or `|vecB_R| = - vecB_A + vecB_B + vecB_C + vecB_D + vecB_E`
` = - vecB_A - vecB_A = - 2 vecB_A`
or `|vecB_R| = 2vecB_A = 2 ( (mu_0 I)/(2pi R) ) = (mu_0 I)/(pi R)`
Perpendicular to OA, towards left .