Ncert Exercises & Solutions

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 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?

Hint: Use the ampere-circuital law.

Step 1: (a) Supposes the five wires A, B, C, D, and E be perpendicular to the plane of paper at locations as shown in the figure.

Thus, magnetic field induction due to five wires will be represented by various sides of a closed pentagon in one order, lying in the plane of the paper. So, its value is zero.

Step 2: (b) Since, the vector sum of the magnetic field produced by each wire at O is equal to 0.

Therefore, magnetic induction produced by one current-carrying wire is equal in
the magnitude of resultant of four wires and opposite in direction.

Therefore, the field if the current in one of the wires (say A) is switched off is

\frac{μ_{0}}{2 π} \frac{l}{R}

$\frac{μ_{0}}{2 π} \frac{l}{R}$ perpendicular to AO towards the left.

Step 3: (c) If the current in wire A is reversed, then total magnetic field induction at O;

Magnetic field induction due to wire A + magnetic field induction due to wires B, C, D, and E

= \frac{μ_{0}}{4 πR} \frac{2 I}{R}

$= \frac{μ_{0}}{4 πR} \frac{2 I}{R}$

(acting perpendicular to AO towards left)

+ \frac{μ_{0}}{2 π} \frac{I}{R}

$+ \frac{μ_{0}}{2 π} \frac{I}{R}$ (acting perpendicular AO towards left)

= \frac{μ_{0} I}{πR}

$= \frac{μ_{0} I}{πR}$ acting perpendicular AO towards left.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions