Question

A horizontal rod .20m long is mounted on a balance and carries a current. At the location of the rod a uniform horizontal magnetic field has magnitude .067 T and direction perpendicular to the rod. That magnetic field has magnitude by the balance and is found to be .13 N. What is the current?

Answer 1

Given that the length of the horizontal rod is \(L=0.200 \mathrm{~m}\)

The magnitude of the magnetic field is \(B=0.067 \mathrm{~T}\)

The magnetic force on the rod is \(F=0.13 \mathrm{~N}\)

Angle between current and magnetic field \(\theta=90.0^{\circ}\)

We know that the force on a current carrying rod due to the magnetic field is given by \(\vec{F}=I \vec{L} \times \vec{B}\)

\(|\vec{F}|=B I L \sin \theta\)

Now the current passes through the rod is \(I=\frac{F}{L B \sin \theta}\)

\(=\frac{(0.13 \mathrm{~N})}{(0.200 \mathrm{~N})(0.067 \mathrm{~T}) \sin 90.0^{\circ}}\)

\(=9.70 \mathrm{~A}\)