A uniform magnetic field B = 4×10⁻³ k̂ T exerts a force F = (4î+3ĵ)×10⁻¹⁰ N on a particle of charge q = 10⁻⁹ C moving in the xy-plane. What is the velocity of the particle?

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For a charged particle moving in a magnetic field, the magnetic force is $\mathbf{F} = q(\mathbf{v}\times\mathbf{B})$ . If $\mathbf{v} = v_x\hat{i}+v_y\hat{j}$ and $\mathbf{B}=B\hat{k}$ , then $\mathbf{v}\times\mathbf{B} = v_y B\hat{i} - v_x B\hat{j}$ , which lets you solve for $v_x$ and $v_y$ by matching components of the given force.

A uniform magnetic field $\mathbf{B} = 4\times10^{-3}\,\hat{k}\,T$ exerts a force $\mathbf{F} = (4\hat{i}+3\hat{j})\times10^{-10}\,N$ on a particle of charge $q = 10^{-9}\,C$ moving in the $xy$ -plane. What is the velocity of the particle?

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← Two concentric semicircular arcs of radii R₁ and R₂ (with R₁<R₂) are joined by straight radial segments to form a closed loop carrying current I, as shown in a typical textbook figure. What is the magnitude of the magnetic field at the common centre O?An infinitely long conductor PQR is bent to form a right angle, with current I flowing through the path PQ and QR. The magnetic field due to this current at a point M (located perpendicular to PQ at the bend) is H₁. Now another infinitely long straight conductor QS is connected at Q (collinear with PQ, extending in the opposite direction), so that the current becomes I/2 in QR as well as in QS, while the current in PQ remains unchanged at I. The magnetic field at M is now H₂. What is the ratio H₁:H₂? →

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