A. A particle with mass 1.71×103 kg and charge of +1.62×108 C has, at a given in

Question

A. A particle with mass 1.71×103 kg and charge of +1.62×108 C has, at a given instant, a velocity of 3.50×104 m/s along the +y-axis, as shown in the figure.

What is the magnitude of the particle’s acceleration produced by a magnetic field of magnitude 1.23 T in the xy plane, directed at an angle of 45.0 counterclockwise from the +x-axis?

B. An insulated circular ring of diameter 6.45 cmcarries a I2 = 12.0 A current and is tangent to a very long, straight insulated wire carrying I1 = 7.00 A of current, as shown in

Find the magnitude of the magnetic field at the center of the ring due to this combination of wires.

Explanation / Answer

A)

Mass of the particle = m = 1.71x10^(- 3) kg
Charge of the particle = q = + 1.62x10^(- 8) C
velocity of the particle = v = 3.5x10^4 m/s along the y-axis
Magnetic field = B = 1.23 T. making angle = 45º
with the x-axis in the x-y plane
(i, j, k are unit vectors along x-, y-, and z- axes
respectively.)
Force acting on the particle = F
Vector F = q[vector v x vector B]
= q[(v j) x {(Bx) i + (By) j)}]
= q[- v(Bx) k + vBy*0]
= - qv(Bx) k = - qvB cos 45º k
Acceleration = vector a = F/m = - [(qvB)/(2m)] k = -0.288 k^ m/s^@

- k shows that the direction is along the negative z-axis.

B) B = (uoI2/2r) – (uoI1/2pi*r) = (4pi*10^-7*12/2* 0.03225) – (4pi*10^-7*7/2pi* 0.03225)

        = (2.34*10^-4) – (4.34*10^-5) = 1.906*10^-4 T (outward)

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