A long rigid wire lies along the X - axis and carries a current of 10 A in the positive X - direction. Round the wire, the external magnetic field is B → = i ^ + 2 x 2 j ^ with x in meters and B in Tesla. The magnetic force (in S I units) on the segment of the wire between x = 1 m and x = 4 m is

d F → = I d l → × B → F = I ∫ d x i ^ × i ^ + 2 x 2 j ^ = 10 ∫ x = 1 x = 4 2 x 2 d x k ^ = 20 x 3 3 x = 1 x = 4 = 20 3 4 3 - 1 3 = 420 N d F → = I ( d l → × B → ) F = I ∫ ( d x i ^ ) × ( i ^ + 2 x 2 j ^ ) = 10 ∫ x = 1 x = 4 2 x 2 d x k ^ = 20 [ x 3 3 ] x = 1 x = 4 = 20 3 [ 4 3 - 1 3 ] = 420 N

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Physics - Magnetism Question with Solution | TestHub

PhysicsMagnetismMagnetic torque & energyEasy2 minPYQ_2020

PhysicsEasysingle choice

A long rigid wire lies along the $X$ - axis and carries a current of $10 A$ in the positive $X$ - direction. Round the wire, the external magnetic field is $\vec{B} = \hat{i} + 2 x^{2} \hat{j}$ with $x$ in meters and $B$ in Tesla. The magnetic force (in $S I$ units) on the segment of the wire between $x = 1 m$ and $x = 4 m$ is

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Answer:

Solution:

$d \vec{F} = I (d \vec{l} \times \vec{B})$

$F = I \int (d x \hat{i}) \times (\hat{i} + 2 x^{2} \hat{j})$

$= 10 \int_{x = 1}^{x = 4} 2 x^{2} d x \hat{k}$

$= 20 {[\frac{x^{3}}{3}]}_{x = 1}^{x = 4} = \frac{20}{3} [4^{3} - 1^{3}] = 420 N$

$d \vec{F} = I (d \vec{l} \times \vec{B})$

$F = I \int (d x \hat{i}) \times (\hat{i} + 2 x^{2} \hat{j})$

$= 10 \int_{x = 1}^{x = 4} 2 x^{2} d x \hat{k}$

$= 20 {[\frac{x^{3}}{3}]}_{x = 1}^{x = 4} = \frac{20}{3} [4^{3} - 1^{3}] = 420 N$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:MagnetismSubtopic:Magnetic torque & energy

⏱ 2mℹ️ Source: PYQ_2020

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