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 the figure. Each carries current out of the plane of the paper.
$(a)$ What will be the magnetic induction at a point on the axis $O$? The axis is at a distance $R$ from each wire.
$(b)$ What will be the field if the current in one of the wires (say $A$) is switched off?
$(c)$ What if the current in one of the wires (say $A$) is reversed?
$(a)$ What will be the magnetic induction at a point on the axis $O$? The axis is at a distance $R$ from each wire.
$(b)$ What will be the field if the current in one of the wires (say $A$) is switched off?
$(c)$ What if the current in one of the wires (say $A$) is reversed?
(N/A) As shown in the diagram,all five wires $A, B, C, D, E$ are perpendicular to the plane of the paper and carry current in the same direction (outwards).
Due to the symmetry of the regular pentagon,the magnetic field vectors at the center $O$ due to each wire have equal magnitudes $B = \frac{\mu_{0} I}{2 \pi R}$ and are directed such that their vector sum is zero.
Therefore,the net magnetic field at $O$ is $0$.
$(b)$ Let the magnetic fields due to wires $A, B, C, D, E$ at point $O$ be $\vec{B}_{A}, \vec{B}_{B}, \vec{B}_{C}, \vec{B}_{D}, \vec{B}_{E}$ respectively.
From part $(a)$,$\vec{B}_{A} + \vec{B}_{B} + \vec{B}_{C} + \vec{B}_{D} + \vec{B}_{E} = 0$.
If current in wire $A$ is switched off,$\vec{B}_{A} = 0$.
The resultant field is $\vec{B}_{net} = \vec{B}_{B} + \vec{B}_{C} + \vec{B}_{D} + \vec{B}_{E} = -\vec{B}_{A}$.
The magnitude is $|\vec{B}_{net}| = |\vec{B}_{A}| = \frac{\mu_{0} I}{2 \pi R}$.
The direction is opposite to the direction of $\vec{B}_{A}$,which is perpendicular to $OA$.
$(c)$ If the current in wire $A$ is reversed,its magnetic field becomes $-\vec{B}_{A}$.
The resultant field is $\vec{B}_{R} = -\vec{B}_{A} + \vec{B}_{B} + \vec{B}_{C} + \vec{B}_{D} + \vec{B}_{E}$.
Since $\vec{B}_{A} + \vec{B}_{B} + \vec{B}_{C} + \vec{B}_{D} + \vec{B}_{E} = 0$,we have $\vec{B}_{B} + \vec{B}_{C} + \vec{B}_{D} + \vec{B}_{E} = -\vec{B}_{A}$.
Substituting this into the expression for $\vec{B}_{R}$:
$\vec{B}_{R} = -\vec{B}_{A} + (-\vec{B}_{A}) = -2\vec{B}_{A}$.
The magnitude is $|\vec{B}_{R}| = 2 |\vec{B}_{A}| = 2 \left( \frac{\mu_{0} I}{2 \pi R} \right) = \frac{\mu_{0} I}{\pi R}$.
Due to the symmetry of the regular pentagon,the magnetic field vectors at the center $O$ due to each wire have equal magnitudes $B = \frac{\mu_{0} I}{2 \pi R}$ and are directed such that their vector sum is zero.
Therefore,the net magnetic field at $O$ is $0$.
$(b)$ Let the magnetic fields due to wires $A, B, C, D, E$ at point $O$ be $\vec{B}_{A}, \vec{B}_{B}, \vec{B}_{C}, \vec{B}_{D}, \vec{B}_{E}$ respectively.
From part $(a)$,$\vec{B}_{A} + \vec{B}_{B} + \vec{B}_{C} + \vec{B}_{D} + \vec{B}_{E} = 0$.
If current in wire $A$ is switched off,$\vec{B}_{A} = 0$.
The resultant field is $\vec{B}_{net} = \vec{B}_{B} + \vec{B}_{C} + \vec{B}_{D} + \vec{B}_{E} = -\vec{B}_{A}$.
The magnitude is $|\vec{B}_{net}| = |\vec{B}_{A}| = \frac{\mu_{0} I}{2 \pi R}$.
The direction is opposite to the direction of $\vec{B}_{A}$,which is perpendicular to $OA$.
$(c)$ If the current in wire $A$ is reversed,its magnetic field becomes $-\vec{B}_{A}$.
The resultant field is $\vec{B}_{R} = -\vec{B}_{A} + \vec{B}_{B} + \vec{B}_{C} + \vec{B}_{D} + \vec{B}_{E}$.
Since $\vec{B}_{A} + \vec{B}_{B} + \vec{B}_{C} + \vec{B}_{D} + \vec{B}_{E} = 0$,we have $\vec{B}_{B} + \vec{B}_{C} + \vec{B}_{D} + \vec{B}_{E} = -\vec{B}_{A}$.
Substituting this into the expression for $\vec{B}_{R}$:
$\vec{B}_{R} = -\vec{B}_{A} + (-\vec{B}_{A}) = -2\vec{B}_{A}$.
The magnitude is $|\vec{B}_{R}| = 2 |\vec{B}_{A}| = 2 \left( \frac{\mu_{0} I}{2 \pi R} \right) = \frac{\mu_{0} I}{\pi R}$.
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