Question

In: Physics

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 f

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?

Solutions

Expert Solution

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

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

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


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