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Potential Energy of a Dipole in Uniform Electric Field

A conducting ...

Question

A conducting wire is bent in the form of a parabola $y^{2} = x$ carrying a current $i = 1 A$ as shown in the figure. This wire is placed in a magnetic field $\to B = - 2^k$ T. The unit vector in the direction of force (on the given portion a to b) is :

\frac{3^i + 4^j}{5}

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\frac{^i +^j}{\sqrt{2}}

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\frac{^i + 2^j}{\sqrt{5}}

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\frac{^i - 2^j}{\sqrt{5}}

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Solution

The correct option is A $\frac{^i +^j}{\sqrt{2}}$
Co-ordinates of the point $a : (1, 1)$
Co-ordinates of the point $b : (4, - 2)$
Hence $\to L = (4 - 1)^i + (- 2 - 1)^j = 3^i - 3^j$
$\to F = i (\to L \times \to B) = 1 (3^i - 3^j) \times (- 2^k) = 6^j + 6^i$
$∣ ∣ \to F ∣ ∣ = ∣ ∣ 6^j + 6^i ∣ ∣ = 6 \sqrt{2}$
$^F = \frac{\to F}{∣ ∣ \to F ∣ ∣} = \frac{6^j + 6^i}{6 \sqrt{2}} = \frac{^i +^j}{\sqrt{2}}$

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A conducting wire is bent in the form of a parabola y2=x carrying a current i=1A as shown in the figure. This wire is placed in a magnetic field →B=−2^k T. The unit vector in the direction of force (on the given portion a to b) is :

The correct option is A ^i+^j√2Co-ordinates of the point a:(1,1)Co-ordinates of the point b:(4,−2)Hence →L=(4−1)^i+(−2−1)^j=3^i−3^j→F=i(→L×→B)=1(3^i−3^j)×(−2^k)=6^j+6^i∣∣→F∣∣=∣∣6^j+6^i∣∣=6√2^F=→F∣∣→F∣∣=6^j+6^i6√2=^i+^j√2

A conducting wire is bent in the form of a parabola $y^{2} = x$ carrying a current $i = 1 A$ as shown in the figure. This wire is placed in a magnetic field $\to B = - 2^k$ T. The unit vector in the direction of force (on the given portion a to b) is :