Two similar coils of radius $R$ are lying concentrically with their planes at right angles to each other. The currents flowing in them are $I$ and $2I$,respectively. The resultant magnetic field induction at the centre will be

$A$ current $i$ is flowing in a straight conductor of length $L.$ The magnetic induction at a point on its axis at a distance $\frac{L}{4}$ from its centre will be

The Biot-Savart law for a moving point charge $q$ with velocity $\vec{v}$ is given by $\vec{B} = \frac{\mu_0}{4\pi} \frac{q(\vec{v} \times \vec{r})}{r^3}$. This indicates that the magnetic field $\vec{B}$ produced by an electron moving with velocity $\vec{v}$ is such that:

$A$ horizontal overhead power line is at a height of $4\,m$ from the ground and carries a current of $100\,A$ from east to west. The magnetic field directly below it on the ground is $(\mu _0 = 4\pi \times 10^{-7}\,TmA^{-1})$

$A$ point charge $Q (= 3 \times 10^{-12} \, C)$ rotates uniformly in a vertical circle of radius $R (= 1 \, mm)$. The axis of the circle is aligned along the magnetic axis of the earth. At what value of the angular speed $\omega$,the effective magnetic field at the centre of the circle will be reduced to zero? (Horizontal component of earth's magnetic field is $30 \, \mu T$)

$A$ thin ring of $10 \, cm$ radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of $40 \pi \, rad \, s^{-1}$ about its axis,perpendicular to its plane. If the magnetic field at its centre is $3.8 \times 10^{-9} \, T$,then the charge carried by the ring is close to $\left( \mu_0 = 4 \pi \times 10^{-7} \, N/A^2 \right)$.