Question 11:
moderate
The radius of curvature of the path of the charged particle in a uniform magnetic field is directly proportional to
Question 12:
moderate
Two very long straight parallel wires carry steady currents i and 2i in opposite directions. The
distance between the wires is d. At a certain instant of time a point charge q is at a point
equidistant from the two wires in the plane of the wires. Its instantaneous velocity \(\overrightarrow{v}\)
is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is :
Question 13:
moderate
If a charged particle goes unaccelerated in a region containing electric & magnetic fields:
(a) \(\vec{E}\) must be perpendicular to \(\vec{B}\)
(b) \(\vec{v}\) must be perpendicular to \(\vec{E}\)
(c) \(\vec{v}\) must be perpendicular to \(\vec{B}\)
(d) \(E\) must be equal to \(vB\)
For the net force to be zero, \(\vec{F}_e + \vec{F}_b = 0 ⇒ \vec{E} = -(\vec{v} \times \vec{B})\). Since \(\vec{E}\) is the cross product, it must be perpendicular to both \(\vec{B}\) and \(\vec{v}\).
Question 14:
moderate
In the product \(\vec{F} = q(\vec{v} \times \vec{B}) = q \vec{v} \times (B_x \hat{i} + B_y \hat{j} + B_0 \hat{k})\), for \(q = 1\) and \(\vec{v} = 2\hat{i} + 4\hat{j} + 6\hat{k}\) and \(\vec{F} = 4\hat{i} – 20\hat{j} + 12\hat{k}\). What will be the complete expression for \(vec{B}\)?
Using the relation \(\vec{F} = \vec{v} \times \vec{B}\), we compare vector components: \(4\hat{i} - 20\hat{j} + 12\hat{k} = (4 B_0 - 6 B_y)\hat{i} - (2 B_0 - 6 B_x)\hat{j} + (2 B_y - 4 B_x)\hat{k}\). Testing the values in Option C gives \(B_x = -6\, B_y = -6\, B_0 = -8\), which completely satisfies all equations.
Question 15:
moderate
A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron
The magnetic force on the electron is \(F_m = q(\vec{v} \times \vec{B}) = 0\) since they are parallel. The electric force is \(F_e = -eE\), which acts opposite to its velocity, causing the electron to decelerate and decrease its speed.