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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
Solution
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.
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