Question

The magnetic force per unit length acting on a wire carrying a current of \(4\sqrt{3}\) A and making an angle of \(60^\circ\) with the direction of a uniform magnetic field of 200 mT is:

• A. \(1.8 \, {N/m}\)
• B. \(2.4 \, {N/m}\)
• C. \(0.6 \, {N/m}\)
• D. \(1.2 \, {N/m}\)

Hint:

Magnetic force per unit length on a wire is \( B I \sin \theta \). Convert mT to T and use sine of the angle between current and magnetic field.

Answer 1

The Correct Option is D

The magnetic force per unit length \( F/L \) on a current-carrying wire in a magnetic field is given by: \[ \frac{F}{L} = B I \sin \theta \] where - \( B = 200 \, {mT} = 200 \times 10^{-3} \, {T} = 0.2 \, {T} \), - \( I = 4\sqrt{3} \, {A} \), - \( \theta = 60^\circ \). 
Calculate the force per unit length: \[ \frac{F}{L} = 0.2 \times 4\sqrt{3} \times \sin 60^\circ \] We know \[ \sin 60^\circ = \frac{\sqrt{3}}{2} \] So, \[ \frac{F}{L} = 0.2 \times 4\sqrt{3} \times \frac{\sqrt{3}}{2} = 0.2 \times 4 \times \frac{3}{2} = 0.2 \times 6 = 1.2 \, {N/m} \]