Magnetism and Magnetic Effects: Exam-Pattern Drill (2)

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Tags Magnetism

Question

A long straight wire carries a current I=5I = 5 A. A circular loop of radius r=0.1r = 0.1 m carrying I2=2I_2 = 2 A is placed in the same plane as the wire, with its centre at d=0.2d = 0.2 m from the wire. Find the magnetic field at the centre of the loop due to (a) the long wire, (b) the loop itself, and (c) the net field at the loop’s centre.

Solution — Step by Step

For a long straight wire, B=μ0I2πdB = \frac{\mu_0 I}{2\pi d}. With μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7} T·m/A:

B1=4π×10−7×52π×0.2=5×10−6 TB_1 = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.2} = 5 \times 10^{-6} \text{ T}

For a circular loop, B2=μ0I22rB_2 = \frac{\mu_0 I_2}{2 r}:

B2=4π×10−7×22×0.1=4π×10−6≈1.26×10−5 TB_2 = \frac{4\pi \times 10^{-7} \times 2}{2 \times 0.1} = 4\pi \times 10^{-6} \approx 1.26 \times 10^{-5} \text{ T}

Use the right-hand rule. Take the wire current going up the page; at the loop centre (to the right of the wire), B1B_1 points into the page. Pick the loop current as counterclockwise; B2B_2 points out of the page.

Both fields are along the same axis (perpendicular to the plane). Subtract:

Bnet=B2−B1=1.26×10−5−0.5×10−5=0.76×10−5 TB_{\text{net}} = B_2 - B_1 = 1.26 \times 10^{-5} - 0.5 \times 10^{-5} = 0.76 \times 10^{-5} \text{ T}

Direction: out of the page.

Final answer: B1=5×10−6B_1 = 5 \times 10^{-6} T (into page), B2≈1.26×10−5B_2 \approx 1.26 \times 10^{-5} T (out of page), Bnet≈7.6×10−6B_{\text{net}} \approx 7.6 \times 10^{-6} T (out of page).

Why This Works

Magnetic fields obey vector superposition. Once you’ve identified the direction of each contribution using the right-hand rule, addition becomes a one-dimensional problem. The trick is to be ruthless about directions before plugging in numbers.

The two formulae here — μ0I/(2πd)\mu_0 I / (2\pi d) for a long wire and μ0I/(2r)\mu_0 I / (2r) for a loop centre — appear in roughly 40%40\% of NEET magnetism questions. Memorise them in the order they appear in NCERT.

Alternative Method

Compute each field as a vector B⃗\vec{B} with sign convention “out of page = positive”. Then B1=−5×10−6B_1 = -5 \times 10^{-6} T and B2=+1.26×10−5B_2 = +1.26 \times 10^{-5} T. Sum: +7.6×10−6+7.6 \times 10^{-6} T. Positive means out of the page.

JEE often combines a wire and a loop in the same plane to test direction handling. NEET prefers parallel wires. Both reduce to vector addition in 1D once directions are set.

Common Mistake

Using d=rd = r in the wire formula. Here d=0.2d = 0.2 m is the distance from the wire to the loop centre, while r=0.1r = 0.1 m is the loop radius. They are independent. Sketch the geometry first.

Direction errors dominate this topic. Always re-do the right-hand rule on a fresh diagram before adding fields. A small wrong sign costs the entire answer.

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