Question 347: Electromagnetic Induction

Question

A conductor of length 0.6 m is moving with a speed of $5 \mathrm{~m} / \mathrm{s}$ perpendicular to a magnetic field of intensity $0.8 \mathrm{~Wb} / \mathrm{m}^2$. The induced emf across the conductor is

Accepted Answer

Motional emf is determined by using following formula.
$$ e=B v l \sin 	heta $$
Given, length of conductor, $I=0.6 \mathrm{~m}$
speed, $v=5 \mathrm{~m} / \mathrm{s}$
magnetic field, $B=0.8 \mathrm{~Wb} / \mathrm{m}^2$
Induced emf,
$$ \begin{aligned} e & =B / \mathrm{V} \sin 	heta \quad\left[	heta=90^{\circ} 	ext { as } B 	ext { is } 1 \mathrm{~V}ight] \ & =0.8 	imes 0.6 	imes 5 	imes \sin 90^{\circ} \ & =2.4 \mathrm{~V} \end{aligned} $$

Answer

2.4 V

Answer

2.52 V

Answer

25.2 V

Answer

5.04 V

Electromagnetic Induction

A conductor of length 0.6 m is moving with a speed of 5 \mathrm{~m} / \mathrm{s} perpendicular to a magnetic field of intensity 0.8 \mathrm{~Wb} / \mathrm{m}^2. The induced emf across the conductor is

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