Question

A magnetic field of 0.080 T is in the y-direction. The velocity of wire segment S has a magnitude of 78 m/s and components of 18 m/s in the x-direction, 24 m/s in the y-direction, and 72 m/s in the z-direction. The segment has length 0.50 m and is parallel to the z-axis as it moves.
Part1) Find the motional emf induced between the ends of the segment.

Part 2) What would the motional emf be if the wire segment was parallel to the y-axis?

Answer 1

Answer:

(I). The motional emf induced between the ends of the segment is $2.88i-0.72k$

(II). The motional emf is zero.

Explanation:

Given that,

Magnetic field = 0.080 T

Velocity of wire segment = 78 m/s

Component in x direction = 18 m/s

Component in y direction = 24 m/s

Component in z direction = 72 m/s

Length = 0.50 m

We need to calculate the motional emf induced between the ends of the segment

Using formula of emf

$\epsilon=(B_{x}\times v_{x})l$

Put the value into he formula

$\epsilon=((0,0.080,0)\times(18,24,72))\times0.50$

$\epsilon=((72\times0.080)i-0j+k(-18\times0.080))\times0.50$

$\epsilon=(5.76i-0j-1.44k)\times0.05$

$\epsilon=2.88i-0.72k$

(II). If the wire segment is parallel to the y-axis then the angle between B and v is zero.

We need to calculate the motional emf

Using formula of emf

$\epsilon =Bv\sin\theta\times l$

Here, $\theta = 0$

$\epsilon =Bv\sin0\times l$

$\epsilon = 0$

Hence, (I). The motional emf induced between the ends of the segment is $2.88i-0.72k$

(II). The motional emf is zero.