Physics: In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103v/m is established between the plates (with no magnetic field present). when you add the magnetic field as described in part c, to what value do you have to adjust its magnitude b0 to observe no deflection? assume that the plates are 6.00 cm long and that the distance between them and the screen is 12.0 cm. express your answer numerically in tesla.

In your experiment, you measure a total deflection

lamontcarter

08.01.2022

$B_0 = 1.69 \times 10^{-4}\ T$

Explanation:

given,

total deflection = 4.12 cm

Electric field = 1.1 ×10³ V/m

plate length = 6 cm

distance between them = 12 cm

using formula

$v_0 = \sqrt{\dfrac{q\epsilon_0d}{ym}(\dfrac{d}{2}+L)}$

q = 1.6 × 10⁻¹⁹ C

m = 9.11 x 10⁻³¹ kg

d = 0.06 m

L = 0.12 m

$v_0 = \sqrt{\dfrac{1.6 \times 10^{-19}\times 1.1 \times 10^{3}\times 0.06}{0.0412\times 9.11 \times 10^{-31} }(\dfrac{0.06}{2}+0.12)}$

v_0 = 6496355.63 m/s

$v_0 = \dfrac{E}{B_0}$

$B_0 = \dfrac{E}{v_0}$

$B_0 = \dfrac{1.1\times 10^{3}}{6496355.63}$

$B_0 = 1.69 \times 10^{-4}\ T$

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tcapele252

07.04.2023

2 min 40 s.

Explanation:

Distance = 800 ft

Speed (walking speed) = 300 ft/min

Speed = distance/time

Time, t = 800/300

= 8/3

= 2 min 40 s.

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