Question

At a distance of \(10 \mathrm{~cm}\) from a long straight wire carrying current, the magnetic field is \(4 \times 10^{-2}\). At the distance of $40 \mathrm{~cm}$, the magnetic field will be Tesla. (a) \(1 \times 10^{-2}\) (b) \(2 \times \overline{10^{-2}}\) (c) \(8 \times 10^{-2}\) (d) \(16 \times 10^{-2}\)

Short answer

The short answer is: (a) \(1 \times 10^{-2}\)

Step-by-step solution

Write down the formula for the magnetic field due to a long straight wire carrying current

The formula for the magnetic field B at a distance r from a long straight wire carrying current I is given by: \[ B = \frac{\mu_0 I}{2 \pi r} \] where \(\mu_0\) is the permeability of free space and has a value of \(4 \pi \times 10^{-7} Tm/A\).

Calculate the current I in the wire

We are given the magnetic field at a distance of 10 cm, which is \(B_1 = 4 \times 10^{-2} T\). Using the formula and solving for I, we get: \[ I = \frac{2 \pi r_1 B_1}{\mu_0} \] where \(r_1 = 10 \times 10^{-2} m\) (converting cm to m). Now, substitute the given values and calculate the current I: \[ I = \frac{2 \pi (10 \times 10^{-2})(4 \times 10^{-2})}{(4 \pi \times 10^{-7})} \] \[ I = 2 \times 10^4 A \]

Calculate the magnetic field at a distance of 40 cm

Now we have the current I in the wire, and we want to find the magnetic field at a distance of 40 cm, which is \(r_2 = 40 \times 10^{-2} m\). Using the formula and substituting the given values, we get: \[ B_2 = \frac{\mu_0 I}{2 \pi r_2} \] \[ B_2 = \frac{4 \pi \times 10^{-7} \times 2 \times 10^4}{2 \pi (40 \times 10^{-2})} \] \[ B_2 = 1 \times 10^{-2} T \] So the magnetic field at a distance of 40 cm from the wire is 1 × 10⁻² T. The correct answer is: (a) \(1 \times 10^{-2}\)