Two solenoids acting as short bar magnets P and Q are arranged such that their centres are on the X-axis and are separated by a large distance . The magnetic axes of P and Q are along X and Y-axes, respectively. At a point R, midway between their centres , if B is the magnitude of induction due to Q , then the magnitude of total induction at R due to the both magnets is

Options

A. 3B
B. `sqrt(5)B`
C. `(sqrt(5))/(2)B`
D. `B`

Explanation

To solve the problem, we will analyze the magnetic induction at point R due to the two solenoids P and Q, which act as short bar magnets.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have two solenoids (P and Q) acting as bar magnets.
- The magnetic axis of solenoid P is along the X-axis, and that of solenoid Q is along the Y-axis.
- The centers of P and Q are on the X-axis and separated by a large distance.

2. **Position of Point R**:
- Point R is located midway between the centers of P and Q.
- If the distance between the centers of P and Q is L, then the distance from each solenoid to point R is \( \frac{L}{2} \).

3. **Magnetic Field Due to Solenoid Q**:
- The magnetic field \( B_Q \) at point R due to solenoid Q (which is aligned along the Y-axis) is given as B.
- The direction of \( B_Q \) is along the negative Y-axis.

4. **Magnetic Field Due to Solenoid P**:
- The magnetic field \( B_P \) at point R due to solenoid P (which is aligned along the X-axis) can be calculated using the formula for the magnetic field of a dipole:
\[ B_P = \frac{\mu_0}{4\pi} \frac{2M}{(L/2)^3} = \frac{16\mu_0 M}{\pi L^3} \]
- The direction of \( B_P \) is along the positive X-axis.

5. **Calculating the Total Magnetic Field at Point R**:
- Since the magnetic fields \( B_P \) and \( B_Q \) are perpendicular to each other (90 degrees apart), we can use the Pythagorean theorem to find the resultant magnetic field \( B_{net} \):
\[ B_{net} = \sqrt{B_P^2 + B_Q^2} \]
- Substituting \( B_Q = B \) and \( B_P = 4B \) (since \( B_P \) is calculated to be four times \( B \)):
\[ B_{net} = \sqrt{(4B)^2 + B^2} = \sqrt{16B^2 + B^2} = \sqrt{17B^2} = \sqrt{17}B \]

6. **Final Result**:
- The magnitude of the total induction at point R due to both magnets is:
\[ B_{net} = \sqrt{17}B \]