In J.J. Thomson\'s experiment, what is the vertical deflection of the electron\'

Question

In J.J. Thomson's experiment, what is the vertical deflection of the electron's beam given the crossed fields are E = 1 times 10^5 V/m and B = 0.01 T with the capacitor's plate at 5 cm long? a)3.0 cm b) 22.0 cm c) 8.5 cm d) 5.5 cm A single current loop (of Area A = 0.1 m^2) is placed in a uniform magnetic field as shown, where vector B = (0.2cap i + 0.4 cap j - 0.3 cap k)T A current i = 20 A runs in the loop in the direction shown, if the plane of the loop makes an angle theta = 30 degree with the x-axis, then, the torque of the magnetic field on the loop is: a) - 0.5 cap i - 0.3 cap j - 0.7 cap k, N. m b) 0.6 cap i -0.2 cap j + 1.2 cap k N. m c) 1.5 cap i + 0.7 cap j -0.3 cap k, N. m d) -0.7 cap i - 0.5 cap j-0.3 cap k, N. m

Explanation / Answer

v = E/B = 105/0.01 = 1.e+7 m/s

vertical displacement:

y = (1/2)at2

= (1/2) (qE/m)(x/v)2

= (1/2) (1.6x10-19*105/9.11x10-31)(0.05/1.e+7)2

= 0.22 m

= 22 cm

option b is correct.

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