A straight copper rod of length 0.50\text{ m} and mass 40\text{ g} carries a steady curren… — JAMB UTME

A straight copper rod of length $0.50\text{ m}$ and mass $40\text{ g}$ carries a steady current and is suspended horizontally in a uniform magnetic field of $0.40\text{ T}$ . The field is directed horizontally at an angle of $30^\circ$ to the length of the rod. If the upward magnetic force acting on the rod exactly balances its weight, what is the magnitude of the current flowing through the rod? (Take $g = 10\text{ m/s}^2$ )

A
$0.25\text{ A}$
B
$2.0\text{ A}$
$4.0\text{ A}$ Answer
D
$40\text{ A}$

Answer

The magnitude of the current required to balance the weight of the rod is

4.0\text{ A}

For the rod to remain suspended in equilibrium, the upward magnetic force

F_B = BIL \sin \theta

must balance the downward gravitational force

W = mg

. Substituting

m = 0.040\text{ kg}

g = 10\text{ m/s}^2

B = 0.40\text{ T}

L = 0.50\text{ m}

, and

\theta = 30^\circ

yields

0.40 = 0.40 \times I \times 0.50 \times 0.50

, which solves to

I = 4.0\text{ A}

Step-by-Step Solution

Convert the mass of the rod to kilograms and calculate its weight.

m = 40\text{ g} = 0.040\text{ kg}

, so

W = mg = 0.040\text{ kg} \times 10\text{ m/s}^2 = 0.40\text{ N}

Standard SI units must be used to ensure dimensional consistency.

Express the magnetic force acting on the conductor in terms of current

I

F_B = B I L \sin \theta = (0.40\text{ T}) \times I \times (0.50\text{ m}) \times \sin 30^\circ = 0.10 I\text{ N}

The magnetic force on a current-carrying conductor at an angle

\theta

to a magnetic field is given by

F = BIL \sin \theta

Equate the upward magnetic force to the downward weight to find

I

0.10 I = 0.40 \implies I = \frac{0.40}{0.10} = 4.0\text{ A}

For vertical equilibrium, the net vertical force must equal zero.

Key Concept

Equilibrium of a current-carrying conductor in a uniform magnetic field (

F = BIL \sin \theta = mg

)

Estimated Time:2m 0s

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