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JEE PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

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A solid metal cube of edge length 2 cm is moving in a positive Y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive Z-direction. The potential difference between the two faces of the cube perpendicular to the X-axis is

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Correct option: B

12 mV

Option B (12 mV) is correct because it is the potential difference calculated using the formula V = Blv, where B is the magnetic field, l is the length, and v is the velocity.

More Questions from ELECTRO MAGNETIC INDUCTION

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The current in a long solenoid of radius

R

n

turns per unit length is given by

i=i_0 \sin \omega t

. A coil having

N

turns is wound around it near the centre. Find the induced emf in the coil. The induced emf can be produced in a coil by

A. moving the coil with uniform speed inside
uniform magnetic field

B. moving the coil with non uniform speed inside
uniform magnetic field

C. rotating the coil inside the uniform magnetic
field

D. changing the area of the coil inside the uniform
magnetic field

Choose the correct answer from the options given
below: An inductance coil have the time constant

4 \mathrm{sec}

, if it is cut into two equal parts and connected parallel then new time constant of the circuit- If the coils of a transformer are made up of thick wire, then A loop of area

0.1\,\,{m^2}

rotates with a speed of

\mathrm{60}

rps perpendicular to a magnetic field of

\mathrm{0}\mathrm{.4}\,\,\mathrm{T}

\mathrm{100}

turns in the loop, maximum voltage induced in the loop is A

\mathrm{50}\,\,\mathrm{mH}

coil carries a current of

\mathrm{2 A}

, the energy stored in joule is