A circular coil, carrying current, has radius $R$. The distance from the centre of the coil on the axis where the magnetic induction will be $\frac{1}{27}$ th of its value at the centre of the coil is
There is a ring of radius $r$ having linear charge density $\lambda$ and rotating with a uniform angular velocity $\omega$. The magnitude of the magnetic field produced by this ring at its own centre would be ( $\mu_0=$ permeability of air)
A particle of charge ' $q$ ' and mass ' $m$ ' moves in a circular orbit of radius ' $r$ ' with angular speed ' $\omega$ '. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on
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
WB JEE Subjects
Browse all chapters by subject