Irodov 3.368: A Pure Magnetic Field, Seen Moving – Amit & Deepali Sinha Foundation

Problem Irodov 3.368

In an inertial frame $K$ there is no electric field, only a uniform magnetic field of induction $B = 0.8$ T. A second inertial frame $K'$ moves relative to $K$ with a constant velocity $\vect v$ at the angle $\alpha = 45^\circ$ to $\vect B$ , with speed $v = \beta c$ , $\beta = 0.60$ . In $K'$ , find the magnitude and direction

of the electric field $\vect E'$ ;
of the magnetic field $\vect B'$ .

The uniform field B of the frame K at the angle α to the velocity v of the frame K'; the axes x and x' run along v. — The uniform field $\vect B$ of the frame $K$ at the angle $\alpha$ to the velocity $\vect v$ of the frame $K'$ ; the axes $x$ and $x'$ run along $\vect v$ .

Given

$B = 0.8$ T, $\vect E = 0$ in $K$
$\alpha = 45^\circ$ between $\vect v$ and $\vect B$
$\beta = v/c = 0.60$

Find

(a)

E'

and its direction · (b)

B'

and its direction

Solution

Step 1Split the field along and across the velocity

The relativistic transformation treats the parts of a field along $\vect v$ and across it differently, so resolve $\vect B$ that way:

B_\parallel = B\cos\alpha, \qquad B_\perp = B\sin\alpha \tag{1}

With $\vect E = 0$ in $K$ and $\gamma = 1/\sqrt{1-\beta^2}$ , the transformation of the fields to $K'$ reads

\vect E' = \gamma\,\vect v\times\vect B, \qquad \vect B'_\parallel = \vect B_\parallel, \qquad \vect B'_\perp = \gamma\,\vect B_\perp \tag{2}

Step 2The electric field in K’

The magnitude of $\vect v\times\vect B$ is $vB\sin\alpha$ , so by (2), with $v = \beta c$ ,

E' = \gamma\, vB\sin\alpha = \gamma\beta c B\sin\alpha

Numerically:

E' = \frac{0.60\cdot(3.00\times10^{8})\cdot 0.8\cdot 0.707}{0.80}\ \text{V/m}

E' = \frac{\beta c B\sin\alpha}{\sqrt{1-\beta^2}} \approx 1.3\times10^{8}\text{ V/m}

(3)

Its direction is that of $\vect v\times\vect B$ : normal to the plane of $\vect v$ and $\vect B$ .

Step 3The magnitude and direction of B’

Adding the squares of the two parts of $\vect B'$ from (2), with (1):

B'^2 = B^2\cos^2\alpha + \frac{B^2\sin^2\alpha}{1-\beta^2} = B^2\,\frac{(1-\beta^2)\cos^2\alpha + \sin^2\alpha}{1-\beta^2} = B^2\,\frac{1-\beta^2\cos^2\alpha}{1-\beta^2}

Numerically, with $\cos^2\alpha = 0.50$ :

B' = 0.8\sqrt{\frac{1 - 0.36\cdot 0.50}{1 - 0.36}}\ \text{T}

B' = B\sqrt{\frac{1-\beta^2\cos^2\alpha}{1-\beta^2}} \approx 0.91\text{ T}

(4)

The vector $\vect B'$ lies in the plane of $\vect v$ and $\vect B$ . Its angle $\alpha'$ with $\vect v$ is fixed by the ratio of its two parts:

\tan\alpha' = \frac{B'_\perp}{B'_\parallel} = \frac{B\sin\alpha}{\sqrt{1-\beta^2}\,B\cos\alpha}

\tan\alpha' = \frac{\tan\alpha}{\sqrt{1-\beta^2}}, \quad \alpha' \approx 51^\circ

(5)

So $\vect B'$ is turned away from $\vect v$ , from $45^\circ$ to about $51^\circ$ , on the same side as $\vect B$ , and $\vect E'$ is perpendicular to it.

The fields in K', drawn for β = 0.60: B' keeps the part of B along v and stretches the cross part by γ, so it leans away from v at α'; E' points out of the page. — The fields in $K'$ , drawn for $\beta = 0.60$ : $\vect B'$ keeps the part of $\vect B$ along $\vect v$ and stretches the cross part by $\gamma$ , so it leans away from $\vect v$ at $\alpha'$ ; $\vect E'$ points out of the page.

Check

Units: $cB$ is $(\text{m/s})\cdot\text{T} = \text{V/m}$ , so (3) is a field strength; a form with $B/c$ in place of $cB$ would be in $\text{T}\cdot\text{s/m}$ , not a field strength at all. Invariant: from (3) and (4),

c^2B'^2 - E'^2 = c^2B^2\,\frac{1-\beta^2\cos^2\alpha - \beta^2\sin^2\alpha}{1-\beta^2} = c^2B^2

the same as in $K$ , where $E = 0$ . Limits: at $\beta \to 0$ , $E' \to 0$ and $\vect B' \to \vect B$ ; at $\alpha = 0$ , $E' = 0$ and $B' = B$ .

Remark

The electric field (3) in $K'$ is huge, far beyond what air or any laboratory gap can hold without breakdown, although $K$ contains only a moderate laboratory magnet. Nothing is paradoxical: a charge at rest in $K'$ moves through the magnet’s field in $K$ , and the Lorentz force $q\vect v\times\vect B$ that $K$ assigns to that motion is what $K'$ calls $q\vect E'$ ; the factor $\gamma$ is the relativistic correction to that bookkeeping.