Question

A metal rod with mass 20 gm and length 20 cm is suspended by two identical springs such that initial elongation in the spring is 5 cm. Now a current of 10 A flows through the rod, then it rises by 1 cm to be in equilibrium. The magnitude of magnetic field present in the region is

Answer 1

0.02 T

Answer 2

The problem involves two equilibrium states of a metal rod suspended by springs.

1. Initial Equilibrium (without current):

• Mass of the rod, $m = 20 \, \text{gm} = 0.02 \, \text{kg}$.
• Length of the rod, $L = 20 \, \text{cm} = 0.2 \, \text{m}$.
• Initial elongation of each spring, $x_1 = 5 \, \text{cm} = 0.05 \, \text{m}$.
• Let $k$ be the spring constant of one spring. Since there are two identical springs, the total upward spring force is $2kx_1$.
• The downward force is the gravitational force, $mg$.
• At equilibrium: $$mg = 2kx_1$$ Assuming $g = 10 \, \text{m/s}^2$ (a common approximation in such problems for JEE/NEET unless specified otherwise): $$0.02 \, \text{kg} \times 10 \, \text{m/s}^2 = 2k \times 0.05 \, \text{m}$$ $$0.2 = 0.1k$$ $$k = \frac{0.2}{0.1} = 2 \, \text{N/m}$$

2. Final Equilibrium (with current):

• A current $I = 10 \, \text{A}$ flows through the rod.
• The rod rises by $1 \, \text{cm}$. This means the new elongation of each spring is $x_2 = x_1 - 1 \, \text{cm} = 5 \, \text{cm} - 1 \, \text{cm} = 4 \, \text{cm} = 0.04 \, \text{m}$.
• The magnetic field $B$ is directed into the page (indicated by 'x' marks).
• From the circuit diagram, the current flows from the positive terminal (left) to the negative terminal (right) through the rod. So, the current in the rod is directed to the right.
• Using Fleming's Left-Hand Rule (or $\vec{F} = I(\vec{L} \times \vec{B})$):
  • Direction of magnetic field ( $\vec{B}$ ): Into the page.
  • Direction of current ( $\vec{I}$ ): To the right.
  • Direction of magnetic force ( $\vec{F_B}$ ): Upwards. This upward magnetic force causes the rod to rise.
• The magnitude of the magnetic force is $F_B = BIL\sin\theta$. Since the current is perpendicular to the magnetic field ( $\theta = 90^\circ$, $\sin 90^\circ = 1$), the magnetic force is $F_B = BIL$.
• At the new equilibrium, the upward forces (spring force + magnetic force) balance the downward gravitational force: $$mg = 2kx_2 + BIL$$ Substitute the known values: $$0.02 \, \text{kg} \times 10 \, \text{m/s}^2 = 2 \times 2 \, \text{N/m} \times 0.04 \, \text{m} + B \times 10 \, \text{A} \times 0.2 \, \text{m}$$ $$0.2 = 4 \times 0.04 + 2B$$ $$0.2 = 0.16 + 2B$$ $$2B = 0.2 - 0.16$$ $$2B = 0.04$$ $$B = \frac{0.04}{2}$$ $$B = 0.02 \, \text{T}$$

The magnitude of the magnetic field present in the region is $0.02 \, \text{T}$.