A circular coil of radius $$10$$ cm, $$500$$ turns and resistance $$2$$ Ω is placed with its plane, perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through $$180$$° in $$0.25$$ s. The induced e.m.f. in the coil is (Take$$ $$HE=3.0x$$ $$10-5T):

Question

A circular coil of radius $$10$$ cm, $$500$$ turns and resistance $$2$$ Ω is placed with its plane, perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through $$180$$° in $$0.25$$ s. The induced e.m.f. in the coil is (Take$$ $$HE=3.0x$$ $$10-5T):

Infinity Learn · Accepted Answer

ε$$=$$−dϕ$$Bdt=2NBAt=2$$ × $$500$$×$$3$$ ×$$10$$−$$5$$ × $$3.14$$  ×$$0.1$$  ×$$0.10.25=37.68$$  ×  $$10$$−$$4=3.8$$  ×  $$10$$−$$3$$ V

Motional EMF

Remember concepts with our Masterclasses.

$\begin{array}{l} ε = - \frac{d ϕ_{B}}{d t} \\ = \frac{2 N B A}{t} \\ = \frac{2 \times 500 \times 3 \times 10^{- 5} \times 3.14 \times 0.1 \times 0.1}{0.25} \\ = 37.68 \times 10^{- 4} \\ = 3.8 \times 10^{- 3} V \end{array}$

Ready to Test Your Skills?

free study material

Motional EMF

Remember concepts with our Masterclasses.

A circular coil of radius 10 cm, 500 turns and resistance 2 Ω is placed with its plane, perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. The induced e.m.f. in the coil is (Take HE=3.0x 10-5T):

ε=−dϕBdt=2NBAt=2 × 500×3 ×10−5 × 3.14 ×0.1 ×0.10.25=37.68 × 10−4=3.8 × 10−3 V

Ready to Test Your Skills?

free study material

$\begin{array}{l} ε = - \frac{d ϕ_{B}}{d t} \\ = \frac{2 N B A}{t} \\ = \frac{2 \times 500 \times 3 \times 10^{- 5} \times 3.14 \times 0.1 \times 0.1}{0.25} \\ = 37.68 \times 10^{- 4} \\ = 3.8 \times 10^{- 3} V \end{array}$