In the given figure,the magnetic force on the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L}_{eff} 	imes \vec{B})$,where $\vec{L}_{eff}$ is the vector displace

Vedclass · Accepted Answer

$20 \, N$

Vedclass · Answer

(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L}_{eff} 	imes \vec{B})$,where $\vec{L}_{eff}$ is the vector displacement from the start point $A$ to the end point $C$.The wire consists of a straight segment $AB$ of length $3 \, m$ and a curved segment $BC$ which is a quarter circle of radius $R = 4 \, m$.The coordinates are: $A = (0, 0)$,$B = (3, 0)$,and $C = (3, 4)$.The effective length vector $\vec{L}_{eff} = \vec{AC} = (3 \hat{i} + 4 \hat{j}) \, m$.The magnetic field is directed into the page,so $\vec{B} = -2 \hat{k} \, T$.The force is $\vec{F} = I(\vec{L}_{eff} 	imes \vec{B}) = 2 \, A 	imes [(3 \hat{i} + 4 \hat{j}) 	imes (-2 \hat{k})] \, T$.Using the cross product rules ($\hat{i} 	imes \hat{k} = -\hat{j}$ and $\hat{j} 	imes \hat{k} = \hat{i}$):$\vec{F} = 2 	imes [3(\hat{i} 	imes -2\hat{k}) + 4(\hat{j} 	imes -2\hat{k})] = 2 	imes [-6(-\hat{j}) + -8(\hat{i})] = 2 	imes [6\hat{j} - 8\hat{i}] = (12\hat{j} - 16\hat{i}) \, N$.The magnitude of the force is $F = \sqrt{(12)^2 + (-16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \, N$.

In the given figure,the magnetic force on the wire $ABC$ will be $(B = 2 \, T, I = 2 \, A)$.

Similar Questions

The figure shows an equilateral triangle $ABC$ of side length '$l$' carrying currents as shown,placed in a uniform magnetic field '$B$' perpendicular to the plane of the triangle. The magnitude of the net magnetic force on the triangle is:

Two long and parallel straight wires $A$ and $B$ carrying currents of $10 \ A$ and $4 \ A$ in the same direction are separated by a distance of $2 \ cm$. Estimate the force on a $4 \ cm$ section of wire $A$. ( $\mu_0 = 4 \pi \times 10^{-7} \ SI$ ).

Two parallel wires in free space are $10 \,cm$ apart and each carries a current of $10 \,A$ in the same direction. The force exerted by one wire on the other [per unit length] is

An elastic circular wire of length $l$ carries a current $I$. It is placed in a uniform magnetic field $\vec{B}$ (out of the paper) such that its plane is perpendicular to the direction of $\vec{B}$. The wire will experience:

Vedclass Test Series

Exam Paper Generator

Online Exam Module