A doubly charged ion with velocity 6.4 x 10 +6 m/s moves in a path of radius 30

Question

A doubly charged ion with velocity 6.4 x 10+6 m/s moves in a path of radius 30 cm in a magnetic field of 0.8 T in a mass spectrometer. What is the mass of this ion?

A vertical straight wire carrying an upward 11 A current exerts an attractive force per unit length of 7.2x10-4 N/m on a second parallel wire 7.6 cm away. What is the current that flows through the second wire?

The drawing shows an end-view of two wires. They are long, straight, and perpendicular to the plane of the paper. Their cross sections lie at the corners of a 0.40-m square. The currents in wire 1 and wire 2 are flowing into the page and equal to 13.0 A and 9.50 A respectively. What is the magnitude of the magnetic field at the center of the square?

Explanation / Answer

1.F = BQv for charged particles in electrical fields (Where B is feild strength, Q charge, v velocity)
and seen as it mentions radius, we can assume its moving a complete circle with a resultant force acting inwards with the value F = mv^2/r (Where m is mass, v is velocity, r is radius)

So BQv = mv^2/r
BQ = mv/r
BQr/v = m
Put the radius into meters instead of centimeters, and you should know that a charged ion will carry the value of one unit charge which 1.6 * 10^-19 C, so doubly charged will have twice this.
Place in values... (0.8 * 2 * 1.6 x10^-19 * 30 x10^-2) / (6.4x10^6) = 1.2x10^-26 Kg

2. dF/dl = ?0 i(1) i(2)/ 2?r
7.2 * 10^-4 = 4? * 10^-7 11 i(2) / 2*0.076?
i(2) =7.2* 10^-4 * 0.152?/ 4? * 10^-7 * 11
i(2) = 24.87A (Answer)

3.The magnetic field B =

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