A conductor (shown in the figure) carrying contant current I is kept in the x-y plane in a uniform magnetic field $\vec{B}$. IF F is the magnitude of the total magnetic force acting on the conductor, then the correct statement(s) is (are):

(A) If $\vec{B}$ is along $\hat{z}, F \propto(L+R)$; (B) If $\vec{B}$ is along $\hat{x}, F=0$; (C) If $\vec{B}$ is along $\hat{y}, F \propto(L+R)$

(D) If $\vec{B}$ is along $\hat{z}, F=0$

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Magnetic Effects of Current

A conductor (shown in the figure) carrying contant current I is kept in the x-y plane in a uniform magnetic field

B

. IF F is the magnitude of the total magnetic force acting on the conductor, then the correct statement(s) is (are):

Answer: (A), (B), (C)

For a conductor in a uniform magnetic field, the net force is

F = I (l \times B)

where

l

is the vector from start to end point.

From the figure, the effective length vector is

l = 2 (L + R) \hat{i}

(horizontal displacement from P to Q).

So

F = I {2 (L + R) \hat{i} \times B}

(A) If

B = B \overset{z}{^}

F = 2 I (L + R) B (\hat{i} \times \overset{z}{^}) = - 2 I (L + R) B \hat{j}

, so

F = 2 I B (L + R) \propto (L + R)

. ✓

(B) If

B = B \overset{x}{^}

F = 2 I (L + R) B (\hat{i} \times \hat{i}) = 0

. ✓

(C) If

B = B \overset{y}{^}

F = 2 I (L + R) B (\hat{i} \times \hat{j}) = 2 I (L + R) B \hat{k}

, so

F = 2 I B (L + R) \propto (L + R)

. ✓

(D) is incorrect as shown in (A).

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · One or more correct options