A long wire carries a current of $18 \mathrm{~A} \mathrm{kept}$ along the axis of a long solenoid of radius $1 \mathrm{~cm}$. The field due to the solenoid is $8.0 \times 10^{-3} \mathrm{~T}$. The magnitude of the resultant field at a point $0.6 \mathrm{~mm}$ from the solenoid axis is (Assume, $\mu_0=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}$ )

• A

  $6 \times 10^{-3} \mathrm{~T}$
• B

  $6 \times 10^{-4} \mathrm{~T}$
• C

  $2 \sqrt{7} \times 10^{-3} \mathrm{~T}$
• Correct answer
  D

  $10 \times 10^{-3} \mathrm{~T}$

Solution

Given, current in a long wire, $I=18 \mathrm{~A}$
magnetic field due to a solenoid, $$ \begin{aligned} & \qquad B_1=8 \times 10^{-3} \mathrm{~T} \\ & \text { and } r=0.6 \mathrm{~mm}=0.6 \times 10^{-3} \mathrm{~m} \\ & \text { Radius of solenoid, } R=1 \mathrm{~cm}=1 \times 10^{-2} \mathrm{~m} \\ & \text { Magnetic field due to current carrying wire at a } \\ & \text { point, } P \\ & \qquad \begin{aligned} B_2 & =\frac{\mu_0}{2 \pi} \cdot \frac{I}{r}=2 \times 10^{-7} \times \frac{18}{0.6 \times 10^{-3}} \\ & =6 \times 10^{-3} \mathrm{~T} \quad \text { (in upward direction) } \end{aligned} \end{aligned} $$ 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, $$ \begin{aligned} B & =\sqrt{B_1^2+B_2^2} \\ & =\sqrt{\left(8 \times 10^{-3}\right)^2+\left(6 \times 10^{-3}\right)^2} \\ & =10 \times 10^{-3} \mathrm{~T} \end{aligned} $$

Hence, the correct answer is option: 'D'.