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Furkat [3]
3 years ago
11

A force of 1.5 × 102 N is exerted on a charge of 1.4 × 10–7 C that is traveling at an angle of 75° to a magnetic field.

Physics
2 answers:
hoa [83]3 years ago
8 0

Answer: B

Explanation:

Marysya12 [62]3 years ago
6 0

Option (B) is correct.The magnetic field strength=8.5 x 10² T

Explanation:

the magnetic force Fm is given by

Fm= q V B sinθ

q= charge=1.4 x 10⁻⁷ C

v= velocity= 1.3 x 10⁶ m/s

B= magnetic field strength

Fm= magnetic force= 1.5 x 10² N

θ=75°

so 1.5 x 10²=(1.4 x 10⁻⁷) (1.3 x 10⁶ ) (B) sin75

B=8.5 x 10² T

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Answer:

a) v = 7.67

b) n = 81562 N

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- The height at from which hammer head drops, s12 = 3.00 m

- The amount of distance the I-beam is hammered, s23 = 7.40 cm

- The resistive force by contact of hammer-head and I-beam, F = 60.0 N

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(a) the speed of the hammerhead just as it hits the I-beam and

(b) the average force the hammerhead exerts on the I-beam.

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- We will consider the hammer head as our system and apply the conservation of energy principle because during the journey of hammer-head up till just before it hits the I-beam there are no external forces acting on the system:

                                   ΔK.E = ΔP.E

                                  K_2 - K_1 = P_1- P_2

Where,  K_2: Kinetic energy of hammer head as it hits the I-beam

             K_1: Initial kinetic energy of hammer head ( = 0 ) ... rest

             P_2: Gravitational potential energy of hammer head as it hits the I-beam. (Datum = 0)

             P_1: Initial gravitational potential energy of hammer head      

- The expression simplifies to:

                                K_2 = P_1

Where,                     0.5*m*v2^2 = m*g*s12

                                v2 = √(2*g*s12) = √(2*9.81*3)

                                v2 = 7.67 m/s

- For the complete journey we see that there are fictitious force due to contact between hammer-head and I-beam the system is no longer conserved. All the kinetic energy is used to drive the I-beam down by distance s23. We will apply work energy principle on the system:

                               Wnet = ( P_3 - P_1 ) + W_friction

                               Wnet = m*g*s13 + F*s23

                               n*s23 = m*g*s13 + F*s23

Where,    n: average force the hammerhead exerts on the I-beam.

               s13 = s12 + s23

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                             n = m*g*( s12/s23 + 1) + F

                             n = 200*9.81*(3/0.074 + 1) + 60

                             n = 81562 N

                               

                                                   

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