Question

A metallic rod of mass per unit length 0.5 kg m^-1 is lying horizontally on smooth inclined plane which makes an angle of 30 degree with the horizontal. A magnetic field of strength 0.25 T is acting on it in the vertical direction. When a current I is flowing through it, the rod is not allowed to slide down. the quantity of current required to keep the road stationary is

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

Answer 1

The Correct Option is C

Given: Mass per unit length of the metallic rod is 

Mass per unit length \( \frac{m}{l} = 0.5 \, \text{kg/m} \).

Let \( I \) be the current flowing through the rod. For equilibrium, the forces acting on the rod balance each other:

\( mg \sin 30^\circ = I B \cos 30^\circ \)

Solving for the current \( I \):

\( I = \frac{mg \sin 30^\circ}{B \cos 30^\circ} \)

Substituting the known values:

\( I = \frac{0.5 \times 9.8 \times \sin 30^\circ}{0.25 \times \cos 30^\circ} = 11.32 \, \text{A} \)

Therefore, the current flowing through the metallic rod is \( 11.32 \, \text{A} \).

Answer 2

\( I = \frac{mg}{B} \tan 30^\circ \)

Given Values: 

• Mass \( m = 0.5 \, \text{kg} \)
• Gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \)
• Magnetic field \( B = 0.25 \, \text{T} \)
• Angle \( 30^\circ \)

Substituting the values into the equation:

\( I = \frac{0.5 \times 9.8}{0.25 \times \sqrt{3}} \)

Simplifying the equation:

\( I = \frac{4.9}{0.25 \times 1.732} = \frac{4.9}{0.433} \)

\( I \approx 11.32 \, \text{A} \)

Therefore, the current required for equilibrium is:

\( \boxed{I \approx 11.32 \, \text{A}} \)