Home
Class 12
PHYSICS
A "bar" magnet of moment vec(M)=hat(i)+h...

A "bar" magnet of moment `vec(M)=hat(i)+hat(j)` is placed in a magnetic field induction `vec(B)=3hat(i)+4hat(j)+4hat(k)`.
The torque acting on the magnet is

A

(a)`4 hati-4hatj+hatk`

B

(b)`hati+hatk`

C

(c)`hati-hatj`

D

(d)`hati+hatj+hatk`

Text Solution

AI Generated Solution

The correct Answer is:
To find the torque acting on a bar magnet placed in a magnetic field, we can use the formula for torque (\(\vec{\tau}\)) which is given by the cross product of the magnetic moment (\(\vec{M}\)) and the magnetic field induction (\(\vec{B}\)): \[ \vec{\tau} = \vec{M} \times \vec{B} \] ### Step 1: Identify the vectors The magnetic moment and magnetic field are given as: - \(\vec{M} = \hat{i} + \hat{j}\) - \(\vec{B} = 3\hat{i} + 4\hat{j} + 4\hat{k}\) ### Step 2: Write the vectors in component form We can express these vectors in component form: - \(\vec{M} = (1, 1, 0)\) - \(\vec{B} = (3, 4, 4)\) ### Step 3: Set up the determinant for the cross product The cross product can be calculated using the determinant of a matrix formed by the unit vectors and the components of \(\vec{M}\) and \(\vec{B}\): \[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 3 & 4 & 4 \end{vmatrix} \] ### Step 4: Calculate the determinant To calculate the determinant, we expand it as follows: \[ \vec{\tau} = \hat{i} \begin{vmatrix} 1 & 0 \\ 4 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 0 \\ 3 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 3 & 4 \end{vmatrix} \] Calculating each of the 2x2 determinants: 1. For \(\hat{i}\): \[ \begin{vmatrix} 1 & 0 \\ 4 & 4 \end{vmatrix} = (1)(4) - (0)(4) = 4 \] 2. For \(\hat{j}\): \[ \begin{vmatrix} 1 & 0 \\ 3 & 4 \end{vmatrix} = (1)(4) - (0)(3) = 4 \] 3. For \(\hat{k}\): \[ \begin{vmatrix} 1 & 1 \\ 3 & 4 \end{vmatrix} = (1)(4) - (1)(3) = 4 - 3 = 1 \] ### Step 5: Combine the results Now substituting back into the expression for \(\vec{\tau}\): \[ \vec{\tau} = 4\hat{i} - 4\hat{j} + 1\hat{k} \] Thus, the torque acting on the magnet is: \[ \vec{\tau} = 4\hat{i} - 4\hat{j} + \hat{k} \] ### Final Answer The torque acting on the magnet is: \[ \vec{\tau} = 4\hat{i} - 4\hat{j} + \hat{k} \] ---
Promotional Banner

Similar Questions

Explore conceptually related problems

A magnet of magnetic moment 50 hat(i) A-m^(2) is placed along the x-axis in a magnetic field vec(B)=(0.5hat(i)+3.0hat(j))T . The torque acting on the magnet is

Find the torque of a force vec(F)=-3hat(i)+hat(j)+5hat(k) acting at the point vec(R)=7hat(i)+3hat(j)+hat(k) .

The compenent of vec(A)=hat(i)+hat(j)+5hat(k) perpendicular to vec(B)=3hat(i)+4hat(j) is

Find the torque of the force vec(F)=(2hat(i)-3hat(j)+4hat(k)) N acting at the point vec(r )=(3hat(i)=2hat(j)+3hat(k)) m about the origion.

The torque of force vec(F)=-3hat(i)+hat(j)+5hat(k) acting at the point vec(r)=7hat(i)+3hat(j)+hat(k) is ______________?

A charged particle moves with velocity vec v = a hat i + d hat j in a magnetic field vec B = A hat i + D hat j. The force acting on the particle has magnitude F. Then,

A charged particle moves with velocity vec v = a hat i + d hat j in a magnetic field vec B = A hat i + D hat j. The force acting on the particle has magnitude F. Then,

Find the angle between the vector vec(a) =2 hat(i) + 3hat(j) - 4 hat(k) and vec(b) = 4hat(i) +5 hat(j) - 2hat(k) .

Given two vectors vec(A) = -hat(i) + 2hat(j) - 3hat(k) and vec(B) = 4hat(i) - 2hat(j) + 6hat(k) . The angle made by (A+B) with x-axis is :

If vec(F ) = hat(i) +2 hat(j) + hat(k) and vec(V) = 4hat(i) - hat(j) + 7hat(k) what is vec(F) . vec(v) ?