Online HW 5 Begin Date: 2/16/2018 5:01:00 PM - Due Date: 2/23/2018 4:59:00 PM En

Question

Online HW 5 Begin Date: 2/16/2018 5:01:00 PM - Due Date: 2/23/2018 4:59:00 PM End Date: 2/23/2018 11:59:00 PM (13%) Problem! 8: A cosmic ray proton moving toward the Earth at 4.5x 107 m/s experiences a magnetic force of 1.95 x 10-16 N > What is the strength of the magnetic field if there is a 45° angle between it and the proton's velocity? Grade Summary Deductions 0% Potential 100% cos0 asino atan acotan sinh0) cosh)tan cotanho Submissions Attempts remaining: 6 04% per attempt) detailed view tano sino cotan) acos) Degrees ° Radians Submit Hint Hints:%deduction per hint. Hints remaining: Feedback: deduction per feedback.

Explanation / Answer

here,

the velocity of proton, v = 4.5 * 10^7 m/s

magnetic force , F = 1.95 * 10^-16 N

theta = 45 degree

let the magnitude of magnetic feild be B

F = q * v * B * sin(theta)

1.95 * 10^-16 = 1.6 * 10^-19 * 4.5 * 10^7 * B * sin(45)

solving for B

B = 3.83 * 10^-5 T

the strengthy of the magnetic feild is 3.83 * 10^-5 T

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