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 10^-4 T.

Given: current I = 14 A, radius r = 2.2 cm = 0.022 m, and the wire forms a semicircle so = radians. Find: the magnetic field at the centre O. For a circular arc, the magnetic field at the centre is B = 0 I 4 r For a semicircle, =, so B = 0 I4r Substituting the values, B = (4 10^-7)(14)4(0.022) B = 14 10^-70.022 B = 14 2.2 10^-5 B 20 10^-5 T B = 2 10^-4 T Therefore, the required numerical value is 2.

A straight wire carrying a current of 14 A is bent into a semicir… — Solution | JEE Physics

NVAEasyJEE 2023Biot–Savart Law

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

A straight horizontal wire is bent upward into a semicircular arc with current flowing from left to right, and the centre O is marked below the arc.

Answer

Correct answer:2

Step-by-step solution

Standard Method

Given: current $I = 14 \, \text{A}$ , radius $r = 2.2 \, \text{cm} = 0.022 \, \text{m}$ , and the wire forms a semicircle so $\theta = \pi$ radians.

Find: the magnetic field at the centre $O$ .

For a circular arc, the magnetic field at the centre is

B = \frac{\mu_0 I \theta}{4\pi r}

For a semicircle, $\theta = \pi$ , so

B = \frac{\mu_0 I}{4r}

Substituting the values,

B = \frac{(4\pi \times 10^{-7})(14)}{4(0.022)}

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}

Therefore, the required numerical value is 2.

Common mistakes

Using the formula for a full circle instead of a semicircular arc is incorrect because here the angle is $\pi$ , not $2\pi$ . Use the arc formula with the correct subtended angle.
Not converting the radius from $2.2 \, \text{cm}$ to $0.022 \, \text{m}$ gives a wrong answer because $\mu_0$ is in SI units. Always convert all quantities to SI before substitution.
Including the straight portions of the wire in the field at $O$ is incorrect here because the solution uses only the field due to the semicircular arc. Follow the geometry used in the given working.

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