Question

A long wire carries a current of $18A$ kept along the axis of a long solenoid of radius $1cm$. The field due to the solenoid is $8.0\times 10^{-3}T$. The magnitude of the resultant field at a point $0.6mm$ from the solenoid axis is (Assume, ${\mu }_0=4\pi \times 10^{-7}Tm/A$ )

• A. $6\times 10^{-3}T$
• B. $6\times 10^{-4}T$
• C. $2\sqrt{7}\times 10^{-3}T$
• D. $10\times 10^{-3}T$

Answer 1

Answer: (D)

Given, current in a long wire, $I=18A$

magnetic field due to a solenoid,
$B_1=8\times 10^{-3}T$
and $r=0.6mm=0.6\times 10^{-3}m$
Radius of solenoid, $R=1cm=1\times 10^{-2}m$
Magnetic field due to current carrying wire at a point, $P$
$B_2=\frac{{\mu }_0}{2\pi }\cdot \frac{I}{r}=2\times 10^{-7}\times \frac{18}{0.6\times 10^{-3}}$
$=6\times 10^{-3}T$ (in upward direction)
Since, current carrying wire is placed along the axis of solenoid, hence magnetic field produced by wire and magnetic field due to solenoid at point $P$, both are perpendicular to each other, Hence, the magnitude of the resultant magnetic field,
$B=\sqrt{B_1^2+B_2^2}$
$=\sqrt{{\left(8\times 10^{-3}\right)}^2+{\left(6\times 10^{-3}\right)}^2}$
$=10\times 10^{-3}T$