Question

A long, straight wire carries a current i. The magnetising field intensity H is measured at a point P close to the wire. A long, cylindrical iron rod is brought close to the wire, so that the point P is at the centre of the rod. The value of H at P will ______________ .

Options

• increase many times

• decrease many times

• remain almost constant

• become zero

Answer 1

remain almost constant

 

From the Biot-Savart law, magnetic field (B) at a point P close to the wire carrying current i is given by,

\[\overrightarrow{B}  = \frac{\mu_0}{4\pi}\frac{i \overrightarrow{dl} \times \overrightarrow{r}}{r^3}\]

Magnetising field intensity (H) will be,

\[H = \frac{B}{\mu_0}   =   \frac{1}{4\pi}\frac{i \overrightarrow{dl} \times \overrightarrow{r}}{r^3}\]

Now, as the cylindrical rod is brought close the wire such that centre of the rod is at P, then distance of point P from the wire(r) will remain same. Hence, magnetic field intensity will remain almost constant. Also even when the rod is carrying any current then B will be zero at the centre of the rod so the value of Magnetising field intensity will remain the same at point P.