For magnetic forces on a current-carrying wire, use F = ILB sinθ. Ensure L and B are in consistent units.
The force on a straight conductor in a magnetic field is given by:
\[ F = I L B \sin \theta \]
Here:
Substitute the values:
\[ F = (2)(0.05)(0.75)(1) = 0.075 \, \text{N} \]
The force is given as \(\frac{x}{130} \, \text{N}\). Equating:
\[ \frac{x}{130} = 0.075 \implies x = 0.075 \cdot 130 = 9.75 \]
Thus, the value of \(x\) is 9.

Force on 5cm side is
∣F∣= ILB sinθ
=(2)(5×10−2)×43×1312=1309N
So, x=9
Hence, The correct answer is 9.

Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.
The magnetic field is a field created by moving electric charges. It is a force field that exerts a force on materials such as iron when they are placed in its vicinity. Magnetic fields do not require a medium to propagate; they can even propagate in a vacuum. Magnetic field also referred to as a vector field, describes the magnetic influence on moving electric charges, magnetic materials, and electric currents.