A circular coil with a radius of \(10.0 \mathrm{~cm}\) has 100 turns of wire and carries a current, \(i=100 . \mathrm{mA} .\) It is free to rotate in a region with a constant horizontal magnetic field given by \(\vec{B}=(0.0100 \mathrm{~T}) \hat{x}\). If the unit normal vector to the plane of the coil makes an angle of \(30.0^{\circ}\) with the horizontal, what is the magnitude of the net magnetic torque acting on the coil? 27.61 At \(t=0\) an electron crosses the positive \(y\) -axis (so \(x=0\) ) at \(60.0 \mathrm{~cm}\) from the origin with velocity \(2.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction. It is in a uniform magnetic field. a) Find the magnitude and the direction of the magnetic field that will cause the electron to cross the \(x\) -axis at \(x=60.0 \mathrm{~cm}\). b) What work is done on the electron during this motion? c) How long will the trip take from \(y\) -axis to \(x\) -axis?

Step by step solution

01

Determine the magnetic torque on the circular coil

First, we need to find the area of the circular coil: Area = \(πr^2\), where r = 0.10 m Now, calculate the magnetic field component in the plane of the coil: \(B_{plane} = B \times \cos{θ}\), where θ = 30.0° After determining \(B_{plane}\), find the magnetic flux passing through the coil: Magnetic flux = \(n \times Area \times B_{plane}\), where n = 100 The net magnetic torque on the coil is determined by the formula: Torque = nABisini, B is magnetic field, i is current We already have the \(B_{plane}\), and the current i = 100 mA. Calculate the magnetic torque using the given values.

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