A square shaped coil of area 70 cm^2 having 600 turns rotates in a magnetic field of 0.4 wbm^-2, about an axis which is parallel to one of the side of the coil and perpendicular to the direction of…

Given: Area of coil A = 70 10^-4 m^2, number of turns N = 600, magnetic field B = 0.4 T, rotational speed 500 revolutions per minute, and the plane of the coil is inclined at 60^ with the field. Find: The instantaneous induced emf. First convert rotational speed into angular velocity: = 500 2 60 = 50 3 Using = 227, = 503 227 = 110021 rad/s For a rotating coil, instantaneous emf is E = NAB where is the angle between the area vector and the magnetic field. Here the plane of the coil makes 60^ with the field, so the area vector makes 30^ with the field. Therefore, E = NAB 30^ Substitute the values directly: E = 600 70 10^-4 0.4 50 3 12 Using = 227, E = 600 70 10^-4 0.4 503 227 12 44 V Therefore, the instantaneous emf is 44 V.

A square shaped coil of area 70 cm^2 having 600 turns rotates in… — Solution | JEE Physics

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A square shaped coil of area $70 \, \text{cm}^2$ having $600$ turns rotates in a magnetic field of $0.4 \, \text{wbm}^{-2}$ , about an axis which is parallel to one of the side of the coil and perpendicular to the direction of field. If the coil completes $500$ revolution in a minute, the instantaneous emf when the plane of the coil is inclined at $60^\circ$ with the field, will be _____ V. (Take $\pi = \frac{22}{7}$ )

Answer

Correct answer:44

Step-by-step solution

Standard Method

Given: Area of coil $A = 70 \times 10^{-4} \, \text{m}^2$ , number of turns $N = 600$ , magnetic field $B = 0.4 \, \text{T}$ , rotational speed $500$ revolutions per minute, and the plane of the coil is inclined at $60^\circ$ with the field.

Find: The instantaneous induced emf.

First convert rotational speed into angular velocity:

\omega = \frac{500 \times 2\pi}{60} = \frac{50\pi}{3}

Using $\pi = \frac{22}{7}$ ,

\omega = \frac{50}{3} \times \frac{22}{7} = \frac{1100}{21} \, \text{rad/s}

Angle Interpretation and Calculation

For a rotating coil, instantaneous emf is

E = NAB\omega \sin \theta

where $\theta$ is the angle between the area vector and the magnetic field.

Here the plane of the coil makes $60^\circ$ with the field, so the area vector makes $30^\circ$ with the field. Therefore,

E = NAB\omega \sin 30^\circ

Direct Substitution

Substitute the values directly:

E = 600 \times 70 \times 10^{-4} \times 0.4 \times \frac{50\pi}{3} \times \frac{1}{2}

Using $\pi = \frac{22}{7}$ ,

E = 600 \times 70 \times 10^{-4} \times 0.4 \times \frac{50}{3} \times \frac{22}{7} \times \frac{1}{2} \approx 44 \, \text{V}

Therefore, the instantaneous emf is $44 \, \text{V}$ .

Common mistakes

Using $\sin 60^\circ$ instead of $\sin 30^\circ$ . This is wrong because the emf formula uses the angle between the area vector and the magnetic field, not the angle between the plane of the coil and the field. First convert plane angle to area-vector angle.
Not converting area from $\text{cm}^2$ to $\text{m}^2$ . This gives an answer larger by a factor of $10^4$ . Always write $70 \, \text{cm}^2 = 70 \times 10^{-4} \, \text{m}^2$ before substitution.
Using rotational frequency directly as $500$ instead of converting to angular velocity. The formula needs $\omega$ in $\text{rad/s}$ , so use $\omega = 2\pi f$ with $f = \frac{500}{60} \, \text{s}^{-1}$ .

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