9303753286127016 Using ε=− dt dϕ and ε=iR find the current in the loop after the external field has stopped changing. A dt di =−( μ 0 πa 2R )i B dt di =− μ 0 a R i C dt di = μ 0 a −2R i D dt di = 3μ 0 πa −2R i

Using ε=− dt dϕ and ε=iR find the current in the loop after the external field has stopped changing. A dt di =−( μ 0 πa 2R )i B dt di =− μ 0 a R i C dt di = μ 0 a −2R i D dt di = 3μ 0 πa −2R i

byBalkishan Agrawal -June 17, 2022

Question

Using $ε = - \frac{d ϕ}{d t}$ and $ε = i R$ find the current in the loop after the external field has stopped changing.

$\frac{d i}{d t} = - (\frac{2 R}{μ _{0} π a}) i$

$\frac{d i}{d t} = - \frac{R}{μ _{0} a} i$

$\frac{d i}{d t} = \frac{- 2 R}{μ _{0} a} i$

$\frac{d i}{d t} = \frac{- 2 R}{3 μ _{0} π a} i$

Medium

Solution

Verified by GMS

Correct option is A)

$ε = \frac{- d ϕ}{d t}$

$= \frac{- μ _{0} π a}{2} \frac{d i}{d t} = i R$

or $\frac{d i}{d t} = - (\frac{2 R}{μ _{0} π a}) i$

Solve any question of Electromagnetic Induction with:-

Patterns of problems

Tags: Physics

Using ε=− dt dϕ and ε=iR find the current in the loop after the external field has stopped changing. A dt di =−( μ 0 πa 2R )i B dt di =− μ 0 a R i C dt di = μ 0 a −2R i D dt di = 3μ 0 πa −2R i

Using $ε = - \frac{d ϕ}{d t}$ and $ε = i R$ find the current in the loop after the external field has stopped changing.

$\frac{d i}{d t} = - (\frac{2 R}{μ _{0} π a}) i$

$\frac{d i}{d t} = - \frac{R}{μ _{0} a} i$

$\frac{d i}{d t} = \frac{- 2 R}{μ _{0} a} i$

$\frac{d i}{d t} = \frac{- 2 R}{3 μ _{0} π a} i$

$ε = \frac{- d ϕ}{d t}$

$= \frac{- μ _{0} π a}{2} \frac{d i}{d t} = i R$

or $\frac{d i}{d t} = - (\frac{2 R}{μ _{0} π a}) i$

Post a Comment

Contact Form

Using ε=− dt dϕ ​ and ε=iR find the current in the loop after the external field has stopped changing. A dt di ​ =−( μ 0 ​ πa 2R ​ )i B dt di ​ =− μ 0 ​ a R ​ i C dt di ​ = μ 0 ​ a −2R ​ i D dt di ​ = 3μ 0 ​ πa −2R ​ i

Using ε=−dtdϕ​ and ε=iR find the current in the loop after the external field has stopped changing.

dtdi​=−(μ0​πa2R​)i

dtdi​=−μ0​aR​i

dtdi​=μ0​a−2R​i

dtdi​=3μ0​πa−2R​i

ε=dt−dϕ​=2−μ0​πa​dtdi​=iRor dtdi​=−(μ0​πa2R​)i