Name: Electric Field: Time Varying Magnetic Field
Uploaded: 2019-07-19
Duration: 8 min 32 s
Description: Did you know that the cause of the movement of charges in a stationary circular conducting coil placed in a time-varying magnetic field is due to an induced...

Question

A small coil of area

5 {cm}^{2}

, number of turns

50

is kept normal to a uniform magnetic field

B

tesla. The coil is rotated about its diameter through

180 °

, in

15 ms

. The total charge flow is

50 mC

. Find

B

, assume resistance

R = 10 Ω .

Accepted Answer

We are given, area of coil, $A = 5 {cm}^{2} = 5 \times 10^{- 4} m$ , number of turns, $50 mC$ , resistance, $R = 10 Ω$ and total charge flow, $q = 50 mC = 50 \times 10^{- 3} C$ .

When the coil is rotated about its diameter through $180 °$ , the direction of the magnetic flux passing through it reverses. We know that the magnetic flux is given as, $ϕ = NBAcosθ$ .

At the start, the angle made by the area of the loop and the magnetic field is $θ = 0 °$ .

$∴$ Magnetic flux, $ϕ_{1} = NBAcosθ$ . Putting the value of $θ = 0 °$ we get,

$\Rightarrow ϕ_{1} = NBAcos (0 °) = NBA \times 1 = NBA$ .

When the coil is rotated by $180 °$ , $θ = 180 °$ . Putting the value of $θ = 180 °$ we get,

$\Rightarrow ϕ_{2} = NBAcos (180 °) = NBA \times - 1 = - NBA$

Change in flux is given as, $Δϕ = ϕ_{2} - ϕ_{1}$ . Putting the values we get,

$\Rightarrow Δϕ = - NBA - NBA = - 2 NBA$ .

Let $t$ be the time taken by the flux to change. Then the e.m.f. induced by the change in the magnetic flux is given by,

$ε = - \frac{Δϕ}{t} = - (\frac{- 2 NBA}{t}) = \frac{2 NBA}{t} . . . (i)$

E.M.F can also be written as, $ε = IR$ . Current is equal to the charge $(q)$ flowing per unit time $(t)$ ,

$∴ ε = \frac{qR}{t} . . . . . (ii)$ . From equations $(i)$ and $(ii)$ we can write,

$\Rightarrow \frac{qR}{t} = \frac{2 NBA}{t} \Rightarrow qR = 2 NBA$ $\Rightarrow B = \frac{qR}{2 NA}$ .

Substituting the given values we get,

$\Rightarrow B = \frac{(50 \times 10^{- 3}) \times 10}{2 \times 50 \times (5 \times 10^{- 4})} = \frac{50 \times 10^{- 2}}{100 \times (5 \times 10^{- 4})} = \frac{0.5}{0.05}$

$\Rightarrow B = \frac{0.5}{0.05} = 10 T$ .

A small coil of area $5 {cm}^{2}$ , number of turns $50$ is kept normal to a uniform magnetic field $B$ tesla. The coil is rotated about its diameter through $180 °$ , in $15 ms$ . The total charge flow is $50 mC$ . Find $B$ , assume resistance $R = 10 Ω .$

Important Questions on Electromagnetic Induction

Learn and Prepare for any exam you want !

A small coil of area 5 cm2, number of turns 50 is kept normal to a uniform magnetic field B tesla. The coil is rotated about its diameter through 180°, in 15 ms . The total charge flow is 50mC. Find B, assume resistance R=10Ω.

Important Questions on Electromagnetic Induction

Learn and Prepare for any exam you want !

A small coil of area $5 {cm}^{2}$ , number of turns $50$ is kept normal to a uniform magnetic field $B$ tesla. The coil is rotated about its diameter through $180 °$ , in $15 ms$ . The total charge flow is $50 mC$ . Find $B$ , assume resistance $R = 10 Ω .$