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WBJEEConsider the motion of a positive point charge in a region where there...
Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields \(\vec{E}=E_{0} \hat{j}\) and \(\vec{B}=B_{0} \hat{j}\). At time \(t=0\), this charge has velocity \(\vec{v}\) in the in the \(x\)-y plane, making an angle \(\theta\) with the \(x\)-axis. Which of the following option(s) is (are) correct for time \(t>0\) ?
Solution:
If \(\theta=0^{\circ}\), the charged particle is projected along \(x\)-axis, due to magnetic field, \(\vec{B}\) the charged particle will tend to move in a circular path in \(y-z\) plane but due to force of electric field \(\vec{E}\), the particle will move in a helical path with increasing pitch. Hence options (A) and (B) are wrong. If \(\theta=10^{\circ}\), we can resolve velocity into two rectangular components. One along \(x\)-axis \(\left(v \cos 10^{\circ}\right)\) and one along \(y\)-axis \(\left(v \sin 10^{\circ}\right)\). Due to \(v \cos 10^{\circ}\), the particle will move in circular path and due to \(v \sin 10^{\circ}\) plus the force due to electric field, the particle will undergo helical motion with its pitch increasing.
If \(\theta=90^{\circ}\), the charge is moving along the magnetic field. Therefore the force due to magnetic field is zero. But the force due to electric field will accelerate the particle along \(y\)-axis.
If \(\theta=90^{\circ}\), the charge is moving along the magnetic field. Therefore the force due to magnetic field is zero. But the force due to electric field will accelerate the particle along \(y\)-axis.
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