# A 0.91-m aluminum bar is held with its length parallel to the east-west direction and dropped...

## Question:

A 0.91-m aluminum bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the bar hits the river below, its speed is 18 m/s, and the emf induced across its length is 6.4 x 10{eq}^-4{/eq} V.

Assuming the horizontal component of the Earth's magnetic field at the location of the bar points directly north,

• (a) Determine the magnitude of the horizontal component of the Earth's magnetic field, and
• (b) State the number of which end of the bar is positive: 1 = east, 2 = west.

## Motional emf

When a conductor moves across the magnetic field, an emf is induced which is known as motional electromotive force (emf).

It is given by the formula;

The induced emf {eq}\varepsilon {/eq} is given by;

{eq}\begin{align} \varepsilon **** = Blv \ \ \ \text {(if v and B are perpendiclular)}\\ \end{align} {/eq}

where,

• {eq}B {/eq} is the magnetic field.
• {eq}l {/eq} is the length of a conductor, and
• {eq}v {/eq} is the velocity of a conductor.

Given:

• The length of a aluminum bar is {eq}l = 0.91 \ m {/eq} which is held parallel to the east-west direction.
• The speed of an aluminum bar just before it hits the river is {eq}v= 18 \ m/s {/eq}.
• The emf induced across its length is {eq}\varepsilon= 6.4 \times10^{-4} \ V {/eq}.

Part (a)

Let

• The magnitude of the horizontal component of the earth's magnetic field is {eq}B_h {/eq}.

Since, {eq}v {/eq} and {eq}B_h {/eq} are perpendiclular to each other, so the induced emf is given by the formula;

{eq}\begin{align} \varepsilon & = B_h lv \\ 6.4 \times10^{-4} & = B (0.91)(18) \\ 6.4 \times10^{-4} & = 16.38 B_h \\ B_h & = \dfrac {6.4 \times10^{-4} } { 16.38} \\ \implies B_h & = 3.91 \times10^{-5} \ T \\ \end{align} {/eq}

Hence, the magnitude of the horizontal component of the earth's magnetic field is {eq}3.91 \times10^{-5} \ T {/eq}.

Part (b)

As we know, {eq}v {/eq} and {eq}B_h {/eq} are perpendiclular to each other and we have a formula,

{eq}\begin{align} \varepsilon & = l( \vec{v} \times \vec{B_h} )\\ \varepsilon & = B_h lv \ \ \ \text {(if v and B_h are perpendiclular)}\\ \end{align} {/eq}

To find out the direction of induced emf {eq}\varepsilon {/eq} we use the right-hand rule.

As the aluminum bar is falling downwards, so the {eq}v {/eq} points in a downwards direction.

It is already given in the question that the horizontal component of the earth's magnetic field {eq}B_h {/eq} at the location of the bar points directly north.

Applying the right-hand rule to the vectors, the direction of induced emf {eq}\varepsilon {/eq} is towards east. So, the east end of the aluminum bar is positive

Hence, in this case, option 1 is right.