# A 0.87-m aluminum bar is held with its length parallel to the east-west direction and dropped...

## Question:

A 0.87-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 28 m/s, and the emf induced across its length is {eq}6.7 \times 10^{-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

A conductor when moves across the magnetic field or there is a change in the magnetic flux, 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 & = l( \vec{v} \times \vec{B} )\\ \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 = 0.87 \ 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= 28 \ m/s {/eq}.
• The emf induced across its length is {eq}\varepsilon= 6.7 \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 \\ 6.7 \times10^{-4} & = B (0.87)(28) \\ 6.7 \times10^{-4} & = 24.36B_h \\ B_h & = \dfrac {6.7 \times10^{-4} } { 24.36 } \\ \implies B_h & = 2.75 \times10^{-5} \ T \\ \end{align} {/eq}

Hence, the magnitude of the horizontal component of the earth's magnetic field is {eq}2.75 \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.

Hence, in this case, the option 1 is right.