Find the current through a loop needed to create a maximum torque of 9.00 mn. the loop has 50 square turns that are 15.0 cm on a side and is in a uniform 0.800-t magnetic field.

10 A

Explanation:

τ = Maximum torque of the loop = 9 mN

N = Number of turns in the loop = 50

a = side of the loop = 15 cm = 0.15 m

A = Area of the loop = a² = 0.15² = 0.0225 m²

B = magnitude of magnetic field = 0.800 T

i = magnitude of current in the loop

Maximum torque of the loop is given as

τ = N B i A

Inserting the values

9 = (50) (0.8) (0.0225) i

i = 10 A

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s