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I. Multiple Choice Questions · Q1

Q.The magnetic field at the centre O of the following current loop is: the loop consists of two straight radial segments meeting a semicircular arc of radius rr centred at O -- current II enters along one straight radial lead, traces the semicircular arc, and leaves along the other straight radial lead (both leads point directly at O, and a current element aimed directly at the field point contributes zero field there by the Biot-Savart law).

(a) μ0I4r\dfrac{\mu_0 I}{4r}, into the page
(b) μ0I4r\dfrac{\mu_0 I}{4r}, out of the page
(c) μ0I2r\dfrac{\mu_0 I}{2r}, into the page
(d) μ0I2r\dfrac{\mu_0 I}{2r}, out of the page
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✓ Free question

Step 1. Split the current path into its three pieces: the straight lead in, the semicircular arc of radius rr, and the straight lead out.

Step 2. For each straight lead, every current element I dl⃗I\,d\vec l points directly toward (or away from) the centre O, so the angle θ\theta between dl⃗d\vec l and r^\hat r is 0°0° or 180°180°, and sin⁡θ=0\sin\theta=0. By the Biot-Savart law, dB=μ04πI dlsin⁡θr2=0dB=\dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}=0 for every element of both straight leads -- they contribute nothing to the field at O.

Step 3. The full field at O therefore comes only from the semicircular arc. For a complete circular loop of radius rr, the centre field is Bloop=μ0I/2rB_{loop}=\mu_0I/2r (§3.8.3). A semicircle is exactly half of a full loop, and every element of the semicircle is at the same distance rr from O with θ=90°\theta=90°, so its contribution is exactly half of the full loop's:

B=12×μ0I2r=μ0I4rB = \frac{1}{2}\times\frac{\mu_0 I}{2r} = \frac{\mu_0 I}{4r}

Step 4. Direction: applying the right-hand thumb rule to the sense of current flow around the arc gives a field into the page at O (matching option (a)).

✓Final answer

(a) μ0I4r\dfrac{\mu_0 I}{4r}, into the page

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