The horizontal component of the earth's magnetic field at any place is $$0.36 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^2$$. If the angle of dip at that place is $$60^{\circ}$$, then the value of vertical component of the earth's magnetic field will be (in $$\mathrm{Wb} / \mathrm{m}^2$$ )
Consider the following figure, a uniform magnetic field of 0.2 T is directed along the positive X-axis. The magnetic flux through top surface of the figure.
An ideal coil of $$10 \mathrm{~H}$$ is connected in series with a resistance of $$5 \Omega$$ and a battery of $$5 \mathrm{~V}$$. After $$2 \mathrm{~s}$$, after the connection is made, the current flowing (in ampere) in the circuit is
In the circuit, shown the galvanometer $$G$$ of resistance $$60 \Omega$$ is shunted by a resistance $$r=0.02 \Omega$$. The current through $$R$$ is nearly $$1 \mathrm{~A}$$. The value of resistance $$R$$ (in ohm) is nearly