Answer to Question #208860 in Mechanics | Relativity for Nightmare

Question

A 5m long straight metal rod is kept on a horizontal table along east-west direction.

Assuming earth’s magnetic field as 0.4 G and angle of dip as 600

. Calculate induced emf in

the rod when it is moved with a velocity of 10 m/s along the direction of

(i)horizontal component of earth’s magnetic field

(ii) vertical component of earth’s magnetic field

Answer 1

Explanations & Calculations

Available data

• Magnetic field = $\small 0.4G = 0.4\times10^{-4}\mu T=0.4\times10^{-10}T$
• Angle of dip = $\small 60^0$
• Speed of the moving rod = $\small 10ms^{-1}$
• Length of it = $\small 5m$

Then,

• The horizontal component of the field is

$\qquad\qquad \begin{aligned} \small \vec{B_h}&=\small 0.4G\times\cos60=0.2G \end{aligned}$

• Vertical component of it is

$\qquad\qquad \begin{aligned} \small \vec{B_v}&=\small 0.4G\times\sin 60=0.346G \end{aligned}$

(1) When the rod is moved parallel to the horizontal component, it is the magnetic field's vertical component that is being cut by the rod.

• Then the induced EMF will be

$\qquad\qquad \begin{aligned} \small \vec{E}&=\small |\vec{B_v}|.l.v\\ &=\small 0.346\times10^{-10}T\times5m\times10ms^{-1}\\ &=\small \bold{1.73\times 10^{-9}V} \end{aligned}$

(2) Similarly it is the horizontal component of the field that is being cut when the rod is moved parallel to the vertical component.

• Then the induced EMF will be

$\qquad\qquad \begin{aligned} \small \vec{E}&=\small 0.2\times10^{-10}\times5\times10\\ &=\small \bold{1\,nV} \end{aligned}$