Photon with kinetic energy of 1MeV moves from south to north. It gets an acceleration of 10¹² m/s² by an applied magnetic field (west to east). The value of magnetic field : (Rest mass of proton is 1.6 × 10^–27 kg) :

A particle of mass m and charge q has an initial velocity $$\overrightarrow v = {v_0}\widehat j$$ . If an electric field $$\overrightarrow E = {E_0}\widehat i$$ and magnetic field $$\overrightarrow B = {B_0}\widehat i$$ act on the particle, its speed will double after a time:

An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of magnetic field $$\overrightarrow B = \left( {1.5 \times {{10}^{ - 3}}T} \right)\widehat k$$ at S (See figure). The field extends between x = 0 and x = 2 cm. The electron is detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on the screen) is : (electron’s charge = 1.6 × 10^–19 C, mass of electron = 9.1 × 10^–31 kg)

Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5A. (See figure) ($$\mu $$₀ = 4$$\pi $$ × 10^–7 N-A^–2)