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Moving Charges and Magnetism - Torque on Current Loop and Magnetic Dipole

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A current-carrying loop placed in a uniform magnetic field experiences a torque but no net force. The net force is zero because the forces on opposite sides of the loop are equal in magnitude and opposite in direction.

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The Magnetic Dipole Moment m⃗\vec{m} of a current loop is defined as m⃗=IA⃗\vec{m} = I\vec{A}, where II is the current and A⃗\vec{A} is the area vector. For a coil with NN turns, m⃗=NIA⃗\vec{m} = NI\vec{A}.

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The direction of the magnetic moment m⃗\vec{m} is perpendicular to the plane of the loop, determined by the Right-Hand Thumb Rule (curl fingers in the direction of current, thumb points towards m⃗\vec{m}).

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The torque τ⃗\vec{\tau} acting on the loop is the vector product of the magnetic moment and the magnetic field: τ⃗=m⃗×B⃗\vec{\tau} = \vec{m} \times \vec{B}.

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The magnitude of torque is τ=mBsin⁡θ\tau = mB \sin \theta, where θ\theta is the angle between the magnetic moment (normal to the loop) and the magnetic field lines.

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Stable Equilibrium occurs when θ=0∘\theta = 0^\circ (m⃗\vec{m} is parallel to B⃗\vec{B}); Torque is zero and Potential Energy is minimum. Unstable Equilibrium occurs when θ=180∘\theta = 180^\circ (m⃗\vec{m} is anti-parallel to B⃗\vec{B}); Torque is zero and Potential Energy is maximum.

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The potential energy UU of a magnetic dipole in a uniform magnetic field is given by U=−m⃗⋅B⃗=−mBcos⁡θU = -\vec{m} \cdot \vec{B} = -mB \cos \theta.

📐Formulae

m⃗=NIA⃗\vec{m} = NI\vec{A}

τ⃗=m⃗×B⃗\vec{\tau} = \vec{m} \times \vec{B}

τ=NIABsin⁡θ\tau = NIAB \sin \theta

U=−m⃗⋅B⃗=−mBcos⁡θU = -\vec{m} \cdot \vec{B} = -mB \cos \theta

W=mB(cos⁡θ1−cos⁡θ2)W = mB(\cos \theta_1 - \cos \theta_2)

μl=e2meL\mu_l = \frac{e}{2m_e} L

💡Examples

Problem 1:

A circular coil of 100100 turns and radius 5 cm5 \text{ cm} carries a current of 2 A2 \text{ A}. It is placed in a uniform magnetic field of 0.2 T0.2 \text{ T} such that the plane of the coil makes an angle of 60∘60^\circ with the field. Calculate the magnitude of the torque acting on the coil.

Solution:

  1. Calculate Area AA: A=πr2=π(0.05)2=0.0025π m2A = \pi r^2 = \pi (0.05)^2 = 0.0025\pi \text{ m}^2
  2. Identify the angle θ\theta: The plane makes 60∘60^\circ with BB, so the normal to the plane (vector A⃗\vec{A}) makes θ=90∘−60∘=30∘\theta = 90^\circ - 60^\circ = 30^\circ with BB.
  3. Calculate Torque τ\tau: τ=NIABsin⁡θ\tau = NIAB \sin \theta τ=100×2×(0.0025π)×0.2×sin⁡30∘\tau = 100 \times 2 \times (0.0025\pi) \times 0.2 \times \sin 30^\circ τ=100×2×0.00785×0.2×0.5\tau = 100 \times 2 \times 0.00785 \times 0.2 \times 0.5 τ≈0.157 N m\tau \approx 0.157 \text{ N m}

Explanation:

The torque is calculated using the formula τ=NIABsin⁡θ\tau = NIAB \sin \theta. It is crucial to use the angle between the normal to the loop and the magnetic field, not the angle with the plane itself.

Problem 2:

A magnetic dipole has a magnetic moment of 0.5 A m20.5 \text{ A m}^2. How much work is required to rotate it from its position of stable equilibrium to unstable equilibrium in a uniform magnetic field of 1.2 T1.2 \text{ T}?

Solution:

  1. Stable equilibrium: θ1=0∘\theta_1 = 0^\circ
  2. Unstable equilibrium: θ2=180∘\theta_2 = 180^\circ
  3. Work done W=mB(cos⁡θ1−cos⁡θ2)W = mB(\cos \theta_1 - \cos \theta_2) W=0.5×1.2×(cos⁡0∘−cos⁡180∘)W = 0.5 \times 1.2 \times (\cos 0^\circ - \cos 180^\circ) W=0.6×(1−(−1))W = 0.6 \times (1 - (-1)) W=0.6×2=1.2 JW = 0.6 \times 2 = 1.2 \text{ J}

Explanation:

Work done is the change in potential energy. Moving from θ=0∘\theta = 0^\circ to 180∘180^\circ represents the maximum possible change in potential energy for the dipole.