Question

A straight wire carrying of 14 A is bent into a semi-circular 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 ____ × 10^–4 T.

Answer 1

Correct Answer: 2

The magnetic field due to an arc is given by: $$B = \frac{\mu_0 I \theta}{4 \pi r}$$ where:  $B$ is the magnetic field 
$\mu_0 = 4\pi \times 10^{-7}$ T m/A 
$I = 14$ A 
$\theta = \pi$ radians (semicircle) 
$r = 2.2 \text{ cm} = 0.022 \text{ m}$ 
Plugging in the values: $$B = \frac{(4\pi \times 10^{-7} \text{ T m/A}) (14 \text{ A}) (\pi)}{4 \pi (0.022 \text{ m})}$$ $$B = \frac{14 \pi \times 10^{-7}}{0.022}$$ $$B = \frac{14 \pi}{2.2} \times 10^{-5}$$ $$B \approx 20 \times 10^{-5} \text{ T}$$ $$B = 2 \times 10^{-4} \text{ T}$$ 
Answer: The magnetic field is $2 \times 10^{-4}$ T. Thus, the answer is 2.