CBSE 2025, Set 2

Question

A straight conductor is carrying a current of $2	ext{ A}$ in $+x$ direction along it. A uniform magnetic field $\vec{B} = (0.6\hat{j} + 0.8\hat{k})	ext{ T}$ is switched on, in the region. The force acting on $10	ext{ cm}$ length of the conductor is: A). $(0.12\hat{j} - 0.16\hat{k})	ext{ N}$ B). $(-0.16\hat{j} + 0.12\hat{k})	ext{ N}$ C). $(-0.12\hat{j} + 0.16\hat{k})	ext{ N}$ D). $(0.16\hat{j} - 0.12\hat{k})	ext{ N}$

Edvaya · Accepted Answer

The correct answer is $(-0.16\hat{j} + 0.12\hat{k})	ext{ N}$. The magnetic force $\vec{F}$ acting on a current-carrying conductor is given by the vector cross product:
$$\vec{F} = I(\vec{l} 	imes \vec{B})$$

Given values:
* Current $I = 2	ext{ A}$
* Length vector $\vec{l} = 10	ext{ cm}$ in $+x$ direction $= 0.1	ext{ m } \hat{i}$
* Magnetic field $\vec{B} = 0.6\hat{j} + 0.8\hat{k}	ext{ T}$

Substitute the values into the formula:
$$\vec{F} = 2 \left[ (0.1\hat{i}) 	imes (0.6\hat{j} + 0.8\hat{k}) ight]$$
$$\vec{F} = 2 \left[ 0.06(\hat{i} 	imes \hat{j}) + 0.08(\hat{i} 	imes \hat{k}) ight]$$

Using the properties of cross products of unit vectors ($\hat{i} 	imes \hat{j} = \hat{k}$ and $\hat{i} 	imes \hat{k} = -\hat{j}$):
$$\vec{F} = 2 \left[ 0.06\hat{k} - 0.08\hat{j} ight]$$
$$\vec{F} = -0.16\hat{j} + 0.12\hat{k}$$

$$\boxed{\vec{F} = (-0.16\hat{j} + 0.12\hat{k})	ext{ N}}$$

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Physics > Magnetism Induction And Alternating Current > Moving Charges And Magnetism

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