In the product

$\overrightarrow{\mathrm{F}} =\mathrm{q}(\vec{v} \times \overrightarrow{\mathrm{B}})$

$=\mathrm{q} \vec{v} \times\left(\mathrm{B} \hat{i}+\mathrm{B} \hat{j}+\mathrm{B}_{0} \hat{k}\right)$

For $\mathrm{q}=1$ and $\vec{v}=2 \hat{i}+4 \hat{j}+6 \hat{k}$ and

$\overrightarrow{\mathrm{F}}=4 \hat{i}-20 \hat{j}+12 \hat{k}$

What will be the complete expression for $\vec{B}$ ?

Bob of a simple pendulum of length $l$ is made of iron . The pendulum is oscillating over a horizontal coil carrying direct current. If the time period of the pendulum is $T$ then

A proton is projected with a velocity $10^7\, m/s$, at right angles to a uniform magnetic field of induction $100\, mT$. The time (in second) taken by the proton to traverse $90^o$ arc is  $(m_p = 1.65\times10^{-27}\, kg$ and $q_p = 1.6\times10^{-19}\, C)$

An electron is moving along the positive $X$-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative $X$-axis. This can be done by applying the magnetic field along

A very high magnetic field is applied to a stationary charge. Then the charge experiences