A conducting wire of parabolic shape, initially y=x^2 is moving with velocity V^{\to}=V_0i^{⋀} in a non – uniform magnetic field =\overrightarrow{B}=B_{0}(1+( \frac{y}{L})^{\beta})\overrightarrow{k} , as shown in figure. If V_0.B_0,L and β are positive constants and \\Deltaϕ is the potential difference developed between the ends of the wire, then the correct statement(s) is/are [JEE Adv 2019 P1]

|∆ϕ|= \frac{1}{2}B_{0}V_{0}L for \beta=0

|∆ϕ| is proportional to the length of the wire projected on the y-axis

|∆ϕ| remains the same if the parabola wire is replaced by a straight wire, y=x initially, of length \sqrt2L

|∆\phi|= \frac{4}{3}B_{0}V_{0}L for \beta=2

Electromagnetic Induction

Section: Physics

Question : 8 of 18

Marks: +1, -0

A conducting wire of parabolic shape, initially y=x2 is moving with velocity

→

=V0

⋀

in a non – uniform magnetic field =

→

=B0(1+(‌‌

)β)

→

, as shown in figure. If V0.B0,L and β are positive constants and Δϕ is the potential difference developed between the ends of the wire, then the correct statement(s) is/are

[JEE Adv 2019 P1]

|∆ϕ|=‌‌
1
2
B0V0L for β=0
|∆ϕ| is proportional to the length of the wire projected on the y-axis
|∆ϕ| remains the same if the parabola wire is replaced by a straight wire, y=x initially, of length √2L
|∆φ|=‌‌
4
3
B0V0L for β=2

Validate

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