A circular loop with a radius of 20 cm carries a 10 A current. An electron passes through the centre of the loop, crossing its plane at a speed of 2.0 × 10^6 m s^ -1 . If the electron's direction of motion is at a 30^ ^

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Correct answer: B. 16π × 10^ -19 N

The magnetic field at the centre of a circular loop is given by

B = \dfrac{\mu_0 I}{2R}

Substituting the given values:

B = \dfrac{4\pi \times 10^{-7} \times 10}{2 \times 0.2} = \pi \times 10^{-5} \text{ T}

The direction of this magnetic field is along the axis of the circular loop.

The magnitude of the magnetic force acting on the electron is given by

F = qvB \sin\theta

Substituting the values of charge, velocity, magnetic field, and angle:

F = (1.6 \times 10^{-19}) \times (2.0 \times 10^6) \times (\pi \times 10^{-5}) \times \sin 30^{}^{{}^{\circ}}

F = 1.6 \times 10^{-19} \times 2.0 \times 10^6 \times \pi \times 10^{-5} \times \dfrac{1}{2}

F = 1.6\pi \times 10^{-18} \text{ N} = 16\pi \times 10^{-19} \text{ N}

A circular loop with a radius of 20 cm carries a 10 A current. An...

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