A beam of light of wavelength $$\lambda$$ falls on a metal having work function $$\phi$$ placed in a magnetic field B. The most energetic electrons, perpendicular to the field are bent in circular arcs of radius R. If the experiment is performed for different values of $$\lambda$$, then $$\mathrm{B}^2$$ vs. $$\frac{1}{\lambda}$$ graph will look like (keeping all other quantities constant)
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A charged particle moving with a velocity $$\vec{v}=v_1 \hat{i}+v_2 \hat{j}$$ in a magnetic field $$\vec{B}$$ experiences a force $$\vec{F}=F_1 \hat{i}+F_2 \hat{j}$$. Here $$v_1, v_2, F_1, F_2$$ all are constants. Then $$\overrightarrow{\mathrm{B}}$$ can be
Two straight conducting plates form an angle $$\theta$$ where their ends are joined. A conducting bar in contact with the plates and forming an isosceles triangle with them starts at the vertex at time $$t=0$$ and moves with constant velocity $$\vec{v}$$ to the right as shown in figure. A magnetic field $$\vec{B}$$ points out of the page. The magnitude of emf induced at $$t=1$$ second will be
Three point charges $$\mathrm{q},-2 \mathrm{q}$$ and $$\mathrm{q}$$ are placed along $$x$$ axis at $$x=-{a}, 0$$ and $a$ respectively. As $$\mathrm{a} \rightarrow 0$$ and $$\mathrm{q} \rightarrow \infty$$ while $$\mathrm{q} \mathrm{a}^2=\mathrm{Q}$$ remains finite, the electric field at a point P, at a distance $$x(x \gg a)$$ from $$x=0$$ is $$\overrightarrow{\mathrm{E}}=\frac{\alpha \mathrm{Q}}{4 \pi \epsilon_0 x^\beta} \hat{i}$$. Then