Question

A proton moves perpendicular to a uniform magnetic field at a speed of 1.00 x 107 m/s and
experiences an acceleration of 2.00 x 1013 m/s2 in the positive x-direction when its velocity is
in the positive Z- direction. Determine the magnitude and direction of the field.​

Answer 1

Answer:

Explanation:

allow'slearntothink... more beneficial like allow'sforgethowtoadd utilising the formulation F = (q)(v)(B)(sin ?) we detect that: allow Vector-B, having value "B" T make an attitude ? counterclockwise with i hat a million. (a million.6*10^-19)(a million.18*10^6 m/s)(B)(sin ?) = a million.fifty 3*10^-16 2. (a million.6*10^-19)(2.3*10^6 m/s)(B)(cos ?) = -3.sixty 9*10^-16; remedy for (B)(sin ?) and (B)(cos ?), we get: 3. (B)(sin ?) = a million.fifty 3*10^-16 / (a million.6*10^-19)(a million.18*10^6 m/s) = 8.104*10^-4 4. (B)(cos ?) = -3.sixty 9*10^-16 / (a million.6*10^-19)(2.3*10^6 m/s) = a million.0027*10^-3 squaring and including both consequences then employing trig identity (sin ?)^2 + (cos ?)^2 = a million (B^2)(sin^2 ?) + (B^2)(cos^2 ?) = (B^2)( (sin ?)^2 + (cos ?)^2) = B^2 B^2 = (8.104*10^-4)^2 + (a million.0027*10^-3)^2 = a million.662*10^-6 B = a million.289 mT plug in the fee of B in equation a million or 2 (use arcsin / arccos / sin-a million(x) / cos-a million(x) ) to remedy for ?