A non conducting rod of length 2m is hinged at one end and a charge of 10 1 C is distributed uniformly over it. At t = 0 , it is released from the position shown. There exist a uniform magnetic fiel

Question

A non conducting rod of length 2m is hinged at one end and a charge of $\frac{1}{10} C$ is distributed uniformly over it. At $t = 0$ , it is released from the position shown. There exist a uniform magnetic field of $15 T$ inside the plane of motion of rod. If mass of rod is 100 g, then find the value of hinge force (in N) when rod is rotated by $\frac{π}{2}$ due to gravity.

TestHub · Accepted Answer

Energy conservation from horizontal to vertical (magnetic force does no work): $m g \frac{L}{2} = \frac{1}{2} \cdot \frac{m L ^{2}}{3} ω^{2} \Rightarrow ω^{2} = \frac{3 g}{L} = 15$

Magnetic force on the rotating charged rod (each element moves at v = ωr): $F_{m} = \int_{0}^{L} λω B r d r = \frac{q ω B L}{2} = \frac{0.1 \times 15 \times 15 \times 2}{2} = 1.5 N$ , directed away from the hinge along the rod.

At the vertical position, forces along the rod: hinge force N (towards hinge), weight mg = 1 N and F_m = 1.5 N (away from hinge).

Centripetal need of the centre of mass: $m ω^{2} \frac{L}{2} = 0.1 \times 15 \times 1 = 1.5 N$

$N = m g + F_{m} + m ω^{2} \frac{L}{2} = 1 + 1.5 + 1.5 = 4 N$

Answer: 4

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