Question

A potential difference of 53 mV is developed across the ends of a 12.0-cm-longwire as it moves through a 0.27 T uniform magnetic field at a speed of 5.0 m/s. The magnetic field is perpendicular to the axis of the wire. What is the angle between the magnetic field and the wire’s velocity?

Answer 1

Answer:

The angle between the magnetic field and the wire’s velocity is 19.08 degrees.                                            

Explanation:

Given that,

Potential difference, V = 53 mV

Length of the wire, l = 12 cm = 0.12 m

Magnetic field, B = 0.27 T

Speed of the wire, v = 5 m/s

Due to its motion, an emf is induced in the wire. It is given by :

$\epsilon=Blv\sin\theta$

Here,

$\theta$ is the angle between magnetic field and the wire’s velocity

$\sin\theta=\dfrac{\epsilon}{Blv}\\\\\sin\theta=\dfrac{53\times 10^{-3}}{0.27\times 0.12\times 5}\\\\\sin\theta=0.327\\\\\theta=19.08^{\circ}$

So, the angle between the magnetic field and the wire’s velocity is 19.08 degrees.