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NEET PHYSICSEasy

A cycle wheel of radius 0.5 m0.5 \text{ m} is rotated with a constant angular velocity of 10 rad/s10 \text{ rad/s} in a region of a magnetic field of 0.1 T0.1 \text{ T} which is perpendicular to the plane of the wheel. The EMF generated between its centre and the rim is:

A

0.25 V0.25 \text{ V}

B

0.125 V0.125 \text{ V}

C

0.5 V0.5 \text{ V}

D

zero

Step-by-Step Solution

The induced electromotive force (emf) ε\varepsilon generated between the center (axle) and the rim of a rotating conductor (like a wheel or rod) in a uniform magnetic field perpendicular to the plane of rotation is given by the formula : ε=12BωR2\varepsilon = \frac{1}{2} B \omega R^2 Where: BB is the magnetic field strength (0.1 T0.1 \text{ T}). ω\omega is the angular velocity (10 rad/s10 \text{ rad/s}).

  • RR is the radius or length of the conductor (0.5 m0.5 \text{ m}).

Calculation: ε=12×0.1×10×(0.5)2\varepsilon = \frac{1}{2} \times 0.1 \times 10 \times (0.5)^2 ε=0.5×1×0.25\varepsilon = 0.5 \times 1 \times 0.25 ε=0.125 V\varepsilon = 0.125 \text{ V}

Note: Even if the wheel has multiple spokes, they are connected in parallel between the center and the rim, so the net emf remains the same as that of a single spoke .

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