Magnetism and Magnetic Effects: Application Problems (1)

easy 2 min read

Question

A long straight wire carries a current of I=5I = 5 A. Find the magnitude of the magnetic field at a point r=0.10r = 0.10 m from the wire. Take μ0=4π×107\mu_0 = 4\pi \times 10^{-7} T m/A.

Solution — Step by Step

The field at distance rr from a long straight current-carrying wire:

B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}

B=4π×107×52π×0.10=4π×5×1070.2πB = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.10} = \frac{4\pi \times 5 \times 10^{-7}}{0.2\pi}

The π\pi cancels:

B=4×5×1070.2=20×1070.2=105 TB = \frac{4 \times 5 \times 10^{-7}}{0.2} = \frac{20 \times 10^{-7}}{0.2} = 10^{-5} \text{ T}

Final answer: B=105B = 10^{-5} T =10= 10 µT.

The direction is given by the right-hand rule — curl the fingers in the direction of current, thumb points along the wire; the field circles the wire.

Why This Works

Ampere’s law gives Bdl=μ0Ienc\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}. By symmetry, BB is constant on a circle of radius rr around the wire, so B2πr=μ0IB \cdot 2\pi r = \mu_0 I, giving the formula above.

The factor μ0/(2π)\mu_0/(2\pi) shows up in many wire-related formulas — memorise it as 2×1072 \times 10^{-7} in SI units.

Alternative Method

Use the Biot-Savart law and integrate over the wire — for an infinite straight wire, you’ll recover the same μ0I/(2πr)\mu_0 I / (2\pi r) result. Ampere’s law is faster when symmetry permits.

In NEET, this exact form appears with different numbers nearly every year. JEE Main combines it with force on another current-carrying wire (F/L=μ0I1I2/(2πd)F/L = \mu_0 I_1 I_2 / (2\pi d)). Memorise both forms.

Common Mistake

Students sometimes use B=μ0I/(4πr)B = \mu_0 I / (4\pi r) — that’s the formula for a current element via Biot-Savart, not the full wire. The full infinite wire gives μ0I/(2πr)\mu_0 I / (2\pi r). The factor of 2 difference matters in MCQs.

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