Alight beam in air strikes the surface of water at an angle of 35° to the normal. when it enters the water it bends toward the normal, making an angle of 25.5° to it. the velocity of light in air is approximately 300,000 km/s. what is the approximate velocity of light in water? a. 77,000 km/s b. 225,000 km/s c. 278,000 km/s d. 341,000 km/s

Quick Answer B

Sin(A)/Sin(B) = v1/v2
A = 25.5
B = 35
v2 = 300,000 km/hr
v1 = ???
Note that this is an inverse relationship.
Sin(25.5)/Sin(35) = v1/300000
v1 = [sin(25.5)/sin(35)] * 300000
sin(A) = 0.4305
Sin(B) = 0.5736
v1 = (0.4305)/0.5736 * 300000
v1 = 225000

The magnitude of the force on the wire is 2.68 N.

Explanation:

Given that,

Length of the wire, L = 5 m

Magnetic field, B = 0.37 T

Angle between wire and the magnetic field,

Current in the wire, I = 2.9 A

We need to find the magnitude of the force on the wire. The magnetic force in the wire is given by :

So, the magnitude of the force on the wire is 2.68 N. Hence, this is the required solution.