A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 m s$$^{-1}$$, at right angles to the horizontal component of the earth's magnetic field of $$0.3 \times 10^{-4}$$ Wb m$$^{-2}$$. The value of the induced emf in the wire is:

For a straight conductor of length $$l$$ moving with a velocity $$\vec v$$ in a uniform magnetic field $$\vec B$$, the motional emf is obtained from the expression

$$\mathcal E \;=\;\int (\vec v \times \vec B)\cdot d\vec l.$$

Because the wire is straight and both $$\vec v$$ and $$\vec B$$ are uniform, the integrand is constant, so the integral reduces to a simple dot product:

$$\mathcal E \;=\;(\vec v \times \vec B)\cdot\vec l.$$

Now we interpret each vector in the present situation.

• The horizontal component of the earth’s magnetic field, $$\vec B_h$$, is directed due North, with magnitude $$B_h = 0.3 \times 10^{-4}\;{\rm Wb\,m^{-2}}.$$

• The wire is oriented from North-East to South-West, so the length vector $$\vec l$$ lies along the north-east direction. This direction makes an angle of $$45^\circ$$ with the due-east line. Its magnitude is the length of the wire, $$l = 10\;{\rm m}.$$

• The wire is falling vertically downward with speed $$v = 5\;{\rm m\,s^{-1}}.$$ The downward velocity is at right angles to the horizontal magnetic field, so the cross product simplifies:

$$|\vec v \times \vec B_h| \;=\; v B_h \sin 90^\circ \;=\; vB_h.$$

The vector $$\vec v \times \vec B_h$$ points due East (right-hand rule). Hence the angle between $$\vec v \times \vec B_h$$ (eastward) and $$\vec l$$ (north-east) is $$45^\circ$$. Substituting everything into the dot product, we have

$$\mathcal E \;=\; |\vec v \times \vec B_h|\,|\vec l|\,\cos 45^\circ$$

$$\phantom{\mathcal E} \;=\; (vB_h)\,l \,\cos 45^\circ.$$

Putting the numerical values,

$$vB_h = 5 \times 0.3 \times 10^{-4} = 1.5 \times 10^{-4}\;{\rm V\,m^{-1}}.$$

Multiplying by the length,

$$ (vB_h)l = 1.5 \times 10^{-4} \times 10 = 1.5 \times 10^{-3}\;{\rm V}.$$

Including the factor $$\cos 45^\circ = \dfrac{1}{\sqrt 2}\approx 0.707,$$

$$\mathcal E = 1.5 \times 10^{-3}\, \times 0.707 \;\approx\; 1.06 \times 10^{-3}\;{\rm V}.$$

To one significant figure this is

$$\mathcal E \simeq 1.1 \times 10^{-3}\;{\rm V}.$$

Hence, the correct answer is Option B.