Question

A metallic rod of mass per unit length $0.5 \, kg \, m^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^{\circ}$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction $0.25 \,T$ is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

• A. 11.32 A
• B. 7.14 A
• C. 14.76 A
• D. 5.98 A

Answer 1

The Correct Option is A

To solve this problem, we must determine the current required to keep the rod stationary on an inclined plane using magnetic force. The rod experiences gravitational force due to its mass, and a magnetic force due to the current and the magnetic field.

The gravitational force component that causes the rod to slide down the inclined plane is given by:

F_{\text{gravity}} = mg \sin \theta

where:

• m is the mass per unit length of the rod, 0.5\, \text{kg m}^{-1}
• g is the acceleration due to gravity, approximately 9.8 \, \text{m/s}^2
• \theta = 30^{\circ} is the angle of the inclined plane with the horizontal

Now, the magnetic force (F_{\text{magnetic}}) that keeps the rod stationary is given by:

F_{\text{magnetic}} = I L B \sin \phi

where:

• I is the current flowing through the rod
• L is the length of the rod; however, since we are considering per unit length, L = 1
• B = 0.25 \, T is the magnetic field induction
• \phi = 90^{\circ} is the angle between the magnetic field and the direction of current

For equilibrium (stationary condition) on the inclined plane:

F_{\text{gravity}} = F_{\text{magnetic}}

Substitute the known values:

mg \sin \theta = I B

Solving for I gives:

I = \frac{mg \sin \theta}{B}

Substitute the values:

I = \frac{0.5 \times 9.8 \times \sin 30^{\circ}}{0.25}

Calculate \sin 30^{\circ} = 0.5:

I = \frac{0.5 \times 9.8 \times 0.5}{0.25}

I = \frac{2.45}{0.25} = 9.8 \text{ A}

Therefore, the current needed to keep the rod stationary is approximately 11.32 A.

Thus, the answer is:

The correct option is 11.32 A.