A straight wire carrying a current of 14 A is bent into a semicircular arc of radius 2.2 cm as shown in the figure. The magnetic field produced by the current at the centre O of the arc is _______ $$\times 10^{-4}$$ T

We need to find the magnetic field at the centre O of a semicircular arc carrying a current of 14 A with radius 2.2 cm.

We begin with the Biot-Savart law, which gives the magnetic field at the centre of a circular arc subtending an angle $$\theta$$ (in radians) at the centre: $$B = \frac{\mu_0 I \theta}{4\pi r}$$ where $$\mu_0 = 4\pi \times 10^{-7}$$ T m/A is the permeability of free space, $$I$$ is the current, and $$r$$ is the radius.

For a semicircular arc, $$\theta = \pi$$ radians. Substituting into the formula yields $$B = \frac{\mu_0 I \pi}{4\pi r} = \frac{\mu_0 I}{4r}$$.

Substituting the given values $$I = 14$$ A, $$r = 2.2$$ cm $$= 0.022$$ m, and $$\mu_0 = 4\pi \times 10^{-7}$$ T m/A into this expression gives $$B = \frac{4\pi \times 10^{-7} \times 14}{4 \times 0.022}$$.

To simplify, the numerator is $$4\pi \times 10^{-7} \times 14 = 56\pi \times 10^{-7}$$ and the denominator is $$4 \times 0.022 = 0.088$$. This gives $$B = \frac{56\pi \times 10^{-7}}{0.088} = \frac{56\pi}{0.088} \times 10^{-7}$$.

Next, $$\frac{56}{0.088} = \frac{56000}{88} = \frac{7000}{11} \approx 636.36$$ so $$B \approx 636.36 \times \pi \times 10^{-7} \approx 636.36 \times 3.14 \times 10^{-7}$$ and hence $$B \approx 1998.2 \times 10^{-7} = 2.0 \times 10^{-4} \text{ T}$$.

Converting to the required units confirms that $$B = 2.0 \times 10^{-4}$$ T = $$2 \times 10^{-4}$$ T. Therefore, the magnetic field at the centre of the semicircular arc is 2 $$\times 10^{-4}$$ T.