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Moving Charges and Magnetism - Motion in a Magnetic Field

Grade 12CBSEPhysics

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

🔑Concepts

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A charge qq moving with velocity v\mathbf{v} in a magnetic field B\mathbf{B} experiences a magnetic Lorentz force given by Fm=q(v×B)\mathbf{F}_m = q(\mathbf{v} \times \mathbf{B}).

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The magnitude of the force is F=qvBsin⁡θF = qvB \sin \theta, where θ\theta is the angle between v\mathbf{v} and B\mathbf{B}.

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The magnetic force is always perpendicular to the velocity of the particle; therefore, the work done by the magnetic force is zero, and the kinetic energy of the particle remains constant.

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If v\mathbf{v} is parallel or anti-parallel to B\mathbf{B} (θ=0∘\theta = 0^\circ or 180∘180^\circ), the force is zero and the particle moves in a straight line.

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If v\mathbf{v} is perpendicular to B\mathbf{B} (θ=90∘\theta = 90^\circ), the particle undergoes uniform circular motion. The magnetic force provides the required centripetal force: mv2r=qvB\frac{mv^2}{r} = qvB.

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If the velocity has a component parallel to the field, the particle follows a helical path. The radius depends on the perpendicular component v⊥=vsin⁡θv_{\perp} = v \sin \theta, while the pitch depends on the parallel component v∥=vcos⁡θv_{\parallel} = v \cos \theta.

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The time period TT and frequency ff of the circular motion are independent of the particle's speed and the radius of the orbit.

📐Formulae

F=q(v×B)\mathbf{F} = q(\mathbf{v} \times \mathbf{B})

F=qvBsin⁡θF = qvB \sin \theta

r=mvqBr = \frac{mv}{qB}

T=2πmqBT = \frac{2\pi m}{qB}

f=qB2πmf = \frac{qB}{2\pi m}

Pitch (p)=vcos⁡θ⋅T=2πmvcos⁡θqB\text{Pitch } (p) = v \cos \theta \cdot T = \frac{2\pi m v \cos \theta}{qB}

💡Examples

Problem 1:

An electron moving with a speed of 5×106 m/s5 \times 10^6 \text{ m/s} enters a magnetic field of 0.2 T0.2 \text{ T} at right angles to the field. Calculate the radius of the circular path. (Given: mass of electron m=9.1×10−31 kgm = 9.1 \times 10^{-31} \text{ kg}, charge e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C})

Solution:

r=mvqBr = \frac{mv}{qB} r=9.1×10−31×5×1061.6×10−19×0.2r = \frac{9.1 \times 10^{-31} \times 5 \times 10^6}{1.6 \times 10^{-19} \times 0.2} r=45.5×10−250.32×10−19r = \frac{45.5 \times 10^{-25}}{0.32 \times 10^{-19}} r≈1.42×10−4 mr \approx 1.42 \times 10^{-4} \text{ m}

Explanation:

Since the electron enters perpendicular to the field (θ=90∘\theta = 90^\circ), it moves in a circle. We use the formula for the radius r=mvqBr = \frac{mv}{qB} by substituting the given values for mass, velocity, charge, and magnetic field.

Problem 2:

A proton is accelerated through a potential difference of 10 kV10 \text{ kV} and then enters a uniform magnetic field of 0.5 T0.5 \text{ T} perpendicular to its direction of motion. Find the radius of its path.

Solution:

K.E.=qV=12mv2  ⟹  v=2qVmK.E. = qV = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2qV}{m}} r=mvqB=mqB2qVm=1B2mVqr = \frac{mv}{qB} = \frac{m}{qB}\sqrt{\frac{2qV}{m}} = \frac{1}{B}\sqrt{\frac{2mV}{q}} Substituting values: m=1.67×10−27 kgm = 1.67 \times 10^{-27} \text{ kg}, q=1.6×10−19 Cq = 1.6 \times 10^{-19} \text{ C}, V=104 VV = 10^4 \text{ V}, B=0.5 TB = 0.5 \text{ T} r=10.52×1.67×10−27×1041.6×10−19r = \frac{1}{0.5}\sqrt{\frac{2 \times 1.67 \times 10^{-27} \times 10^4}{1.6 \times 10^{-19}}} r=2×0.20875≈0.0288 m=2.88 cmr = 2 \times \sqrt{0.20875} \approx 0.0288 \text{ m} = 2.88 \text{ cm}

Explanation:

First, the velocity is determined from the kinetic energy gained via the potential difference. Then, that velocity is substituted into the radius formula for circular motion in a magnetic field.