A charged particle enters a uniform magnetic field with a velocity v at right angles to the field. It moves in a circle with period T. An identical particle enters the field with a velocity v/2. What

Physics

Magnetic Field due to current Solutions

v Part A Points 1 and 2 in (Figure 1) are the same distance from the wires as the point where B = 2.2 mT. What is the strength and direction of B at point 1? O (2.2 mT, into the page) O (2.2 mT, out of the page) O (4.4 mT, into the page) O (4.4 mT, out of the page) O The field is zero. Figure < 1 of 1 Submit Request Answer Part B What is the strength and direction of B at point 2? 1. • 2 O (2.2 mT, into the page) 1 • B, out of page O (2.2 mT, out of the page) O (4.4 mT, into the page) O (4.4 mT, out of the page)

Physics

Magnetic Field due to current Solutions

through it, when placed in an external magnetic field B, FB = 11 B sine, o being the angle between I and B 17. The 'cyclotron radius' equals r= mv/qB, the symbols carrying their usual meanings. Question 48 2 pts A straight, flexible length of copper wire is immersed in a magnetic field that is directed into the page. (a) If the wire's current runs in the +x-direction, which way will the wire bend? (b) Which way will the wire bend if the current runs in the -x-direction? Edit View Insert Format Tools Table 12pt Paragraph V B Ave Tv Italic Otv A 11 Aa

Physics

Magnetic Field due to current Solutions

A negative charge is moving with velocity v in an external magnetic field B as shown in the figure. What is the direction of the magnetic force FB acting on the negative charge? # Out of the page O Upward To the right Into the page Downward To the left