Magnetism and Magnetic Effects: Numerical Problems Set (7)

easy • 2 min read
Tags Magnetism

Question

A circular coil of 200200 turns and radius 5 cm5 \text{ cm} carries a current of 2 A2 \text{ A}. Find the magnetic field at its centre. If the coil is placed in an external magnetic field of 0.4 T0.4 \text{ T} such that the plane of the coil is parallel to the external field, find the torque on the coil. Take μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7} T·m/A.

Solution — Step by Step

For a single loop, B=μ0I2RB = \tfrac{\mu_0 I}{2R}. For NN turns:

B=μ0NI2RB = \tfrac{\mu_0 N I}{2R} B=4π×10−7×200×22×0.05B = \tfrac{4\pi \times 10^{-7} \times 200 \times 2}{2 \times 0.05} B=4π×10−7×4000.1=16π×10−4 T≈5.03×10−3 TB = \tfrac{4\pi \times 10^{-7} \times 400}{0.1} = 16\pi \times 10^{-4} \text{ T} \approx 5.03 \times 10^{-3} \text{ T} m=NIA=200×2×π(0.05)2=200×2×π×0.0025m = NIA = 200 \times 2 \times \pi (0.05)^2 = 200 \times 2 \times \pi \times 0.0025 m=π A⋅m2≈3.14 A⋅m2m = \pi \text{ A·m}^2 \approx 3.14 \text{ A·m}^2

When the plane of the coil is parallel to B⃗\vec{B}, the area vector (normal to plane) is perpendicular to B⃗\vec{B}, so θ=90°\theta = 90° between m⃗\vec{m} and B⃗\vec{B}:

τ=mBsin⁡θ=π×0.4×sin⁡90°=0.4π≈1.257 N⋅m\tau = mB\sin\theta = \pi \times 0.4 \times \sin 90° = 0.4\pi \approx 1.257 \text{ N·m}

Final answers: B≈5.03 mTB \approx \mathbf{5.03 \text{ mT}}, τ≈1.26 N⋅m\tau \approx \mathbf{1.26 \text{ N·m}}.

Why This Works

A current loop behaves like a tiny bar magnet with magnetic moment m⃗=NIA⃗\vec{m} = NI\vec{A}, where A⃗\vec{A} points perpendicular to the loop using the right-hand rule. The torque τ⃗=m⃗×B⃗\vec{\tau} = \vec{m} \times \vec{B} is maximum when m⃗⊥B⃗\vec{m} \perp \vec{B} — exactly our case.

The torque tries to rotate the loop until m⃗∥B⃗\vec{m} \parallel \vec{B}, the stable equilibrium.

Alternative Method

Torque can be derived directly from the force on each side of a square (or rectangular) loop. For a circular coil it’s cleaner to use τ=mBsin⁡θ\tau = mB\sin\theta.

Common Mistake

The geometry of “plane parallel to B⃗\vec{B}” trips students. If the plane is parallel to B⃗\vec{B}, the area vector (perpendicular to the plane) makes 90°90° with B⃗\vec{B}, giving maximum torque. Many students wrongly assume “plane parallel” means θ=0\theta = 0, getting zero torque.

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