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.